Lecture Notes 1 Basic Concepts of Mathematics MATH...

Lecture Notes1

Basic Concepts of MathematicsMATH 352

Ivan Avramidi

New Mexico Institute of Mining and TechnologySocorro, NM 87801

June 3, 2004

Author: Ivan Avramidi; File: absmath.tex; Date:June 11, 2007; Time: 15:24

1Textbook: R. J. Bond and W. J. Keane,An Introduction to Abstract Mathematics,Brooks/Cole, 1999

Contents

1 Logic 11.1 LECTURE 1. Statements . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.2 Statements . . . . . . . . . . . . . . . . . . . . . . . . . 21.1.3 Quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . 21.1.4 Negations . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 LECTURE 2. Compound Statements . . . . . . . . . . . . . . . . 51.2.1 Conjunctions and Disjunctions . . . . . . . . . . . . . . . 51.2.2 Truth Tables . . . . . . . . . . . . . . . . . . . . . . . . 51.2.3 Negating Conjunctions and Disjunctions . . . . . . . . . . 61.2.4 Logically Equivalent Statements . . . . . . . . . . . . . . 71.2.5 Tautologies and Contradictions . . . . . . . . . . . . . . . 81.2.6 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3 LECTURE 3. Implications . . . . . . . . . . . . . . . . . . . . . 101.3.1 Truth Table for an Implication . . . . . . . . . . . . . . . 111.3.2 Proving Statements Containing Implications . . . . . . . . 111.3.3 Negating an Implication: Counterexamples . . . . . . . . 121.3.4 Necessary and Sufficient Conditions . . . . . . . . . . . . 131.3.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.4 LECTURE 4. Contrapositive and Converse . . . . . . . . . . . . 151.4.1 Contrapositive . . . . . . . . . . . . . . . . . . . . . . . 151.4.2 Converse . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4.3 Biconditional . . . . . . . . . . . . . . . . . . . . . . . . 161.4.4 Proof by Contradiction . . . . . . . . . . . . . . . . . . . 171.4.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 18

I

II CONTENTS

2 Sets 192.1 LECTURE 5. Sets and Subsets . . . . . . . . . . . . . . . . . . . 19

2.1.1 The Notion of a Set . . . . . . . . . . . . . . . . . . . . . 192.1.2 Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.1.3 Complements . . . . . . . . . . . . . . . . . . . . . . . . 222.1.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.2 LECTURE 6. Combining Sets . . . . . . . . . . . . . . . . . . . 232.2.1 Unions and Intersections . . . . . . . . . . . . . . . . . . 232.2.2 De Morgan’s Laws . . . . . . . . . . . . . . . . . . . . . 242.2.3 Cartesian Products . . . . . . . . . . . . . . . . . . . . . 242.2.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.3 LECTURE 7. Collection of Sets . . . . . . . . . . . . . . . . . . 252.3.1 Power Set . . . . . . . . . . . . . . . . . . . . . . . . . . 252.3.2 Indexing Sets . . . . . . . . . . . . . . . . . . . . . . . . 252.3.3 Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . 272.3.4 The Pigeonhole Principle . . . . . . . . . . . . . . . . . . 272.3.5 Cantor Set . . . . . . . . . . . . . . . . . . . . . . . . . 282.3.6 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 29

3 Functions 313.1 LECTURE 8. Definition and Basic Properties . . . . . . . . . . . 31

3.1.1 Image of a Function . . . . . . . . . . . . . . . . . . . . 323.1.2 Inverse Image . . . . . . . . . . . . . . . . . . . . . . . . 343.1.3 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.2 LECTURE 9. Surjective and Injective Functions. . . . . . . . . . 363.2.1 Surjective Functions . . . . . . . . . . . . . . . . . . . . 363.2.2 Injective Functions . . . . . . . . . . . . . . . . . . . . . 363.2.3 Bijective Functions . . . . . . . . . . . . . . . . . . . . . 373.2.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.3 LECTURE 10. Composition and Inverse Functions . . . . . . . . 383.3.1 Composition of Functions . . . . . . . . . . . . . . . . . 383.3.2 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . 393.3.3 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 40

4 Binary Operations and Relations 414.1 LECTURE 11. Binary Operations . . . . . . . . . . . . . . . . . 41

4.1.1 Associate and Commutative Laws . . . . . . . . . . . . . 414.1.2 Identities . . . . . . . . . . . . . . . . . . . . . . . . . . 42

absmath.tex; June 11, 2007; 15:24; p. 1

CONTENTS III

4.1.3 Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.1.4 Closure . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.1.5 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.1.6 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 44

4.2 LECTURE 12. Equivalence Relations . . . . . . . . . . . . . . . 454.2.1 Relations . . . . . . . . . . . . . . . . . . . . . . . . . . 454.2.2 Properties of Relations . . . . . . . . . . . . . . . . . . . 454.2.3 Equivalence Relations . . . . . . . . . . . . . . . . . . . 464.2.4 Equivalence Classes . . . . . . . . . . . . . . . . . . . . 464.2.5 Partial and Linear Ordering . . . . . . . . . . . . . . . . 484.2.6 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 49

5 The Integers 515.1 LECTURE 13. Axioms and Basic Properties . . . . . . . . . . . . 51

5.1.1 The Axioms of the Integers . . . . . . . . . . . . . . . . 515.1.2 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 535.1.3 The Well-Ordering Principle . . . . . . . . . . . . . . . . 545.1.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 55

5.2 LECTURE 14. Induction . . . . . . . . . . . . . . . . . . . . . . 565.2.1 Induction: A Method of Proof . . . . . . . . . . . . . . . 565.2.2 Other Forms of Induction . . . . . . . . . . . . . . . . . 565.2.3 The Binomial Theorem . . . . . . . . . . . . . . . . . . 585.2.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 59

5.3 LECTURE 15. The Division Algorithm . . . . . . . . . . . . . . 605.3.1 Divisors and Greatest Common Divisors . . . . . . . . . 605.3.2 Euclidean Algorithm . . . . . . . . . . . . . . . . . . . . 625.3.3 Relatively Prime Integers . . . . . . . . . . . . . . . . . 635.3.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 64

5.4 LECTURE 16. Primes and Unique Factorization . . . . . . . . . 655.4.1 Prime Numbers . . . . . . . . . . . . . . . . . . . . . . 655.4.2 Unique Factorization . . . . . . . . . . . . . . . . . . . . 665.4.3 Euclid Theorem . . . . . . . . . . . . . . . . . . . . . . . 685.4.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.5 LECTURE 17. Congruences . . . . . . . . . . . . . . . . . . . . 695.5.1 Congruences and Their Properties . . . . . . . . . . . . . 695.5.2 The Set of Congruence Classes . . . . . . . . . . . . . . 715.5.3 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 75


IV CONTENTS

6 Infinite Sets 776.1 LECTURE 19. Countable Sets . . . . . . . . . . . . . . . . . . . 77

6.1.1 Numerically Equivalent Sets . . . . . . . . . . . . . . . . 776.1.2 Countable Sets . . . . . . . . . . . . . . . . . . . . . . . 786.1.3 Unions of Countable Sets . . . . . . . . . . . . . . . . . 826.1.4 The Rationals are Countable . . . . . . . . . . . . . . . . 836.1.5 Cartesian Products of Countable Sets . . . . . . . . . . . 836.1.6 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 84

6.2 LECTURE 20. Uncountable Sets . . . . . . . . . . . . . . . . . . 856.2.1 Uncountable Sets . . . . . . . . . . . . . . . . . . . . . . 856.2.2 Cantor Theorem . . . . . . . . . . . . . . . . . . . . . . 866.2.3 Continuum Hypothesis . . . . . . . . . . . . . . . . . . . 876.2.4 Schroeder-Bernstein Theorem . . . . . . . . . . . . . . . 886.2.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 92

6.3 LECTURE 21. Collections of Sets . . . . . . . . . . . . . . . . . 936.3.1 Russell’s Paradox . . . . . . . . . . . . . . . . . . . . . 936.3.2 Countable Unions of Countable Sets . . . . . . . . . . . 936.3.3 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 95

7 The Real and Complex Numbers 977.1 LECTURE 22. Fields . . . . . . . . . . . . . . . . . . . . . . . . 97

7.1.1 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 977.1.2 Elementary Properties of Fields . . . . . . . . . . . . . . 987.1.3 Ordered Fields . . . . . . . . . . . . . . . . . . . . . . . 997.1.4 Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . 1017.1.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 103

7.2 LECTURE 23. The Real Numbers . . . . . . . . . . . . . . . . . 1047.2.1 Bounded Sets . . . . . . . . . . . . . . . . . . . . . . . . 1047.2.2 Least Upper Bound and Greatest Lower Bound . . . . . . 1047.2.3 The Archimedean Principle . . . . . . . . . . . . . . . . 1057.2.4 Incompleteness of Rationals . . . . . . . . . . . . . . . . 1067.2.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 107

7.3 LECTURE 24. The Complex Numbers . . . . . . . . . . . . . . . 1087.3.1 Complex Numbers . . . . . . . . . . . . . . . . . . . . . 1087.3.2 Conjugation and Absolute Value . . . . . . . . . . . . . 1097.3.3 Solutions of Equations . . . . . . . . . . . . . . . . . . . 1107.3.4 Polar Form . . . . . . . . . . . . . . . . . . . . . . . . . 1107.3.5 Complex Roots . . . . . . . . . . . . . . . . . . . . . . . 111


CONTENTS V

7.3.6 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 112

Bibliography 113


VI CONTENTS


Chapter 1

Logic

1.1 LECTURE 1. Statements

1.1.1 Proof

Mathematics is an attempt to determine which statements are trueand which are not.

• Inductive reasoning

• Deductive reasoning

• Conjecture is a not verified statement

• Proof is the verification of a statement.

• Axioms are statements that are accepted as given.

• Propositionsare the logical deductions from the axioms.

• Theoremsare particularly important propositions.

• Proof is the logic, the arguments, used to make deductions

• Example

• An integerm is amultiple of an integern if m = kn for some integerk.

1

2 CHAPTER 1. LOGIC

1.1.2 Statements

• Definition 1.1.1 A statementis any declarative sentence that is eithertrue or false.

• Example

• A variable is a symbol that stands for an undetermined number.

•Definition 1.1.2 An open sentenceis any declarative sentence con-taining one or more variables that is not a statement but becomes astatement when the variables are assigned values.

• Notation:P(x), P(x, y)

1.1.3 Quantifiers

• A universal quantifier is a phrase

“for every”, “for all”, etc

• Notation: ∀• Example

• An existential quantifier is a phrase

“there is”, “there exists”, etc

• Notation: ∃• Example

• A bound variable is a variable to which a quantifier is applied.

• A variable that is not bound is afree variable.

• Notation: The symbol

3 is read “such that”

• Remark: The order in which quantifiers appear in a statement is important.

• Example


1.1. LECTURE 1. STATEMENTS 3

1.1.4 Negations

• Example

• Definition 1.1.3 The negation of a statementP is the statement “P isfalse”.

• Notation: The negation ofP is denoted by the symbol

¬P, read “notP”

• Alternative ways to express

¬P: “P is not true”,

or

“It is not true thatP”

• Remark. Exactly one ofP and¬P is true; the other is false.

• Example

• Negation of a statement with a universal quantifier

• Example

• Negation of a statement with an existential quantifier

• Example

• Basic Rulesof negating statements with quantifiers


4 CHAPTER 1. LOGIC

– The negation of the statement

“For all x, P(x)”

is the statement

“For somex, ¬P(x)”

¬(∀x,P(x))⇔ ∃x 3 ¬P(x)

– The negation of the statement

“For somex, P(x)”

is the statement

“For all x, ¬P(x)”

¬(∃x, 3 P(x))⇔ ∀x,¬P(x)

• Example¬ (∀x,∃y 3 P(x, y))⇔ ∃x 3 ∀y,¬P(x, y)

• Definition 1.1.4 (Archimedean Principle)For every real numberx,there is an integern such thatn > x.

1.1.5 Homework

• Read “Introduction for the Student” and to the Chapter 1; pp. xv-xix, 1.

• Reading: Sect 1.1

• Exercises: 1.1[9,11,D5]


1.2. LECTURE 2. COMPOUND STATEMENTS 5

1.2 LECTURE 2. Compound Statements

1.2.1 Conjunctions and Disjunctions

• Compound statementsare statements built up from two or more state-ments.

•Definition 1.2.1 Theconjunction of a statementP and a statementQis the statement

“Both P andQ are true”.

• Notation:

P∧ Q, read “P andQ”

•Definition 1.2.2 Thedisjunction of a statementP and a statementQis the statement

“ P is true orQ is true”.

• Notation:

P∨ Q, read “P or Q”

• Remarks. Similarly for open sentences,P(x) ∨ Q(x) etc.

• Example

1.2.2 Truth Tables

• Statement formsare expressions of the form

P∧ Q, P∨ Q, or ¬P,

whereP andQ are variables representing unspecified statements.

• Statement forms are not statements but they become statements when thevariablesP andQ are specified, that is, replaced by statements.

• Truth tables are tables of truth values of the statement forms.


6 CHAPTER 1. LOGIC

• Example

P Q P∧ QT T TT F FF T FF F F

P Q P∨ QT T TT F TF T TF F F

P ¬PT FF T

• Remarks

• Example

1.2.3 Negating Conjunctions and Disjunctions

• Example

P Q ¬P ¬Q P∧ Q P∨ QT T F F T TT F F T F TF T T F F TF F T T F F

P Q ¬(P∧ Q) ¬(P∨ Q) ¬P∨ ¬Q ¬P∧ ¬QT T F F F FT F T F T FF T T F T FF F T T T T

• That is,

¬(P∧ Q)⇔ ¬P∨ ¬Q

¬(P∨ Q)⇔ ¬P∧ ¬Q

• These statement forms “mean the same thing”.

• Example


1.2.4 Logically Equivalent Statements

• Statements and statement forms arelogically equivalent if they are ex-pressed in different ways but mean the same thing.

• Definition 1.2.3 Two statements arelogically equivalent or justequivalent if they are both true or both false.

•Definition 1.2.4 Two statement forms arelogically equivalent if thesubstitution of statements for the variables in the forms always yieldslogically equivalent statements.

• Notation.

P⇔ Q read “P is equivalent toQ”

• Remark. If two statement forms have the same truth tables then they arelogically equivalent.

• Examples.

1. ¬(P∧ Q)⇔ ¬P∨ ¬Q

2. ¬(P∨ Q)⇔ ¬P∧ ¬Q

3. ¬(∀x,P(x)⇔ ∃x 3 (¬P(x))

4. ¬(∃x 3 P(x)⇔ ∀x, (¬P(x))

5. ¬(∀x, (P(x) ∨ Q(x))⇔ ∃x 3 ((¬P(x)) ∧ (¬Q(x)))

6. ¬(∀x, (P(x) ∧ Q(x))⇔ ∃x 3 ((¬P(x)) ∨ (¬Q(x)))

7. ¬(∃x 3 (P(x) ∨ Q(x))⇔ ∀x, ((¬P(x)) ∧ (¬Q(x)))

8. ¬(∃x 3 (P(x) ∧ Q(x))⇔ ∀x, ((¬P(x)) ∨ (¬Q(x)))

• Examples.

• The statementsFor all x,P(x) or Q(x)

andFor all x,P(x) or for all x,Q(x)

arenot equivalent, that is

∀x, (P(x) ∨ Q(x)) < ∀x,P(x) ∨ ∀x,Q(x)

8 CHAPTER 1. LOGIC

• Proposition.

∃x 3 (P(x) ∨ Q(x))⇔ (∃x 3 P(x)) ∨ (∃x 3 Q(x))) .

Proof: Truth table.

• Example

• One can construct compound statements and statement forms from three ormore statements

• Proposition.(P∧ Q) ∨ (P∧ R)⇔ P∧ (Q∨ R)

Proof:

P Q R P∧ Q P∧ R (P∧ Q) ∨ (P∧ R) Q∨ R P∧ (Q∨ R)T T T T T T T TT T F F T F T TT F T F T T T TT F F F F F F FF T T F F F T FF T F F F F T FF F T F F F T FF F F F F F F F

1.2.5 Tautologies and Contradictions

• Tautology is a statement form that is always true no matter what statementsare substituted for the variables.

• Remark. Each of the truth table values of a tautology is true.

• Example. P∨ ¬P is always true.

• Contradiction is a statement form that is always false no matter what state-ments are substituted for the variables.

• Remark. Each of the truth table values of a contradiction is false.

• Example. P∧ ¬P is always false.

• Negation of a tautology is a contradiction and vice versa.



1.2.6 Homework


• Exercises: 1.2[4,11,12]


10 CHAPTER 1. LOGIC

1.3 LECTURE 3. Implications

• Theorems follow logically from prior propositions

• Propositions follow from axioms

• Axioms are statements that are taken without proof and that serve as thestarting point

• Implications are statements of the form

“If . . . , then . . . ”,

or

“For all . . . , if . . . , then . . . ”.

• Example

• Thepremiseor theassumptionof the statement is the “if” part.

• Theconclusionis the “then” part.

•

Definition 1.3.1 Let P andQ be statements. Theimplication

P⇒ Q

is the statement

“If P is true, thenQ is true”.

• Notation.

P⇒ Q (read “P impliesQ”)

• Example


1.3. LECTURE 3. IMPLICATIONS 11

1.3.1 Truth Table for an Implication

• In the implicationP⇒ Q, P might be unrelated toQ at all.

• P does not have to have causedQ

• The implicationP ⇒ Q means that in all circumstances under whichP istrue,Q is also true.

• In other words, implicationP ⇒ Q means that wheneverP “happens”,Qalso “happens”.

• The implicationP⇒ Q is false only in the case ifP is true andQ is false.

• The implicationP⇒ Q cannot be false ifP is false, even ifQ is false.

• A false statement implies anything!

• Truth values of the statement formP⇒ Q:

P Q P⇒ QT T TT F FF T TF F T

• Examples.

1.3.2 Proving Statements Containing Implications

• Let P(x) andQ(x) be open sentences.

• The expressionP(x) ⇒ Q(x) is an open sentence that becomes a statementwhen the variablex is assigned a valuea.

• Note that the expression∀x,P(x)⇒ Q(x) is a statement.

• We want to prove the statement∀x,P(x)⇒ Q(x).

• Proof: Assume that if the variablex is assigned the valuea, thenP(a) is trueand proceed to proveQ(a).

• P(a) is thehypothesis


12 CHAPTER 1. LOGIC

• Q(a) is theconclusion

• Example

• Remark. Let P(n) be an open sentence. Once we say, “Letn be an integer”,thenP(n) is a statement.

• Example

1.3.3 Negating an Implication: Counterexamples

• In order to prove that a statement containing an implication is false it isnecessary to prove that its negation is true.

• Example.

• The only way for an implication∀x,P(x)⇒ Q(x) to be false is that there isan assigned valuea of the variablex such thatP(a) is true andQ(a) is false.

• A value of a variable served to disprove a statement with a universal quan-tifier is acounterexample.

• Proposition. If P andQ are statement forms, then

¬(P⇒ Q)⇔ P∧ ¬Q

• Proof: Find truth tables.

• If P(x) andQ(x) are open sentences, the negation of the statement

“For all x, P(x)⇒ Q(x)”

is the statement

“There existsx such thatP(x) ∧ ¬Q(x)”,

or

“For somex, P(x) is true andQ(x) is false.”


1.3. LECTURE 3. IMPLICATIONS 13

• Formally¬(∀x,P(x)⇒ Q(x))⇔ ∃x 3 (P(x) ∧ ¬Q(x))

• The value assigned to the variablex that makesP(x) true andQ(x) false isacounterexampleto the statement

“For all x, P(x)⇒ Q(x)”.

• Negation of an implication isnot an implication!

1.3.4 Necessary and Sufficient Conditions

• In the implicationP ⇒ Q (if P is true, thenQ is true) P is a sufficientcondition for Q.

• In order forQ to be true it issufficientthatP be true.

• If P ⇒ Q is true, (it is true that ifP is true thenQ is true), thenQ is anecessary conditionfor P.

• Q mustbe true in order forP to be true.

• In other words, ifQ is false, thenP is false.

• Proposition. The statement

¬Q⇒ ¬P

is equivalent to the statement

P⇒ Q.

(¬Q⇒ ¬P)⇔ (P⇒ Q)

• Proof: Truth table.�

• Remarks.

• If P⇒ Q is true then it isnot necessarythatP be true in order forQ to betrue.


14 CHAPTER 1. LOGIC

• Also, Q is not a sufficient condition forP. That is, even ifQ is true,P maybe false.

• Example

1.3.5 Homework


• Exercises: 1.3[3,13,15]


1.4. LECTURE 4. CONTRAPOSITIVE AND CONVERSE 15

1.4 LECTURE 4. Contrapositive and Converse

1.4.1 Contrapositive

• The statements

P⇒ Q

and

¬Q⇒ ¬P

are logically equivalent

(P⇒ Q)⇔ (¬Q⇒ ¬P)

• Check the truth tables

• To prove the statementP⇒ Q is the same as to verify the statement¬Q⇒¬P

• This means: wheneverQ is false, thenP is false.

• In other words: ifP is true, thenQ cannot be false and, therefore,Q is true.

• Example

•

Definition 1.4.1 Let P andQ be two statements. The statement

¬Q⇒ ¬P

is thecontrapositiveof the statement

P⇒ Q.

• Example


16 CHAPTER 1. LOGIC

1.4.2 Converse

• Example

Definition 1.4.2 The statement

Q⇒ P

is theconverseof the statement

P⇒ Q.

• Example

• The statementsP⇒ Q and its converseQ⇒ P are not necessarily logicallyequivalent!

(P⇒ Q) < (Q⇒ P)

• If the implicationP⇒ Q is true, its converseQ⇒ P could be false.

• Example

1.4.3 Biconditional

• If the statementP ⇒ Q and its converseQ ⇒ P are both true, then thestatementsP andQ are either both true or both false.

• ThenP is a necessary and sufficient condition forQ.

• Thus,P andQ are logically equivalent.

•

Definition 1.4.3 The statement

P⇔ Q

is the statement

(P⇒ Q) ∧ (Q⇒ P).

The symbol⇔ is called thebiconditional.

• Notation. The statement

P⇔ Q is read “P if and only if Q”.

1.4. LECTURE 4. CONTRAPOSITIVE AND CONVERSE 17

• The word “iff” is the shorthand for the phrase “if and only if”.

• Theorem 1.4.1 Let n be an integer. Thenn is even if and only ifn2 iseven.

•

Theorem 1.4.2 Letn be an integer. The following are equivalent state-ments:

1. n is even.

2. n2 is even.

Proof:

• Proposition. The statements

(P⇔ Q)

and

(¬P⇔ ¬Q)

are logically equivalent.

(P⇔ Q)⇔ (¬P⇔ ¬Q)

Proof: By examining the truth tables.

• Corollary 1.4.1 Let n be an integer. Thenn is odd if and only ifn2 isodd.

Proof:

• Example

1.4.4 Proof by Contradiction

• In order to prove an implicationP ⇒ Q one can prove its contrapositive¬Q⇒ ¬P.

• Proof by contradiction.


18 CHAPTER 1. LOGIC

• Proving

¬Q⇒ ¬P

means that we

assume thatQ is false, and prove thatP is false,

getting acontradictionthatP and¬P cannot both be true.

• To prove a statementP by contradiction means to assume thatP is false,and deduce a false statementQ from it.

•

Theorem 1.4.3 Let P be a statement andQ be a false statement. Thenthe statement

¬P⇒ Q

is logically equivalent toP.

Proof:

• Example

1.4.5 Homework


• Exercises: 1.4[2,17,21]


Chapter 2

Sets

2.1 LECTURE 5. Sets and Subsets

2.1.1 The Notion of a Set

• Theset is a basic undetermined concept which cannot be defined formally.

• The set is a collection of objects called theelements.

• Notation.

x ∈ A (readx is in A).

• Example

• Notation.

– N = Z+ the set of positive integers (natural numbers),

– Z the set of integers,

– Q = {mn |m,n ∈ Z,n , 0} the set of rational numbers numbers,

– R+ the set of positive real numbers,

– R the set of real numbers,

– C the set of complex numbers.

• Example

19

20 CHAPTER 2. SETS

• Description of the set of all elements of the setU such that the open sentenceP(x) is true statement

A = {x ∈ U | P(x)}So,A is thetruth set of the open sentenceP(x).

• The setU is a fixed set called auniversal set.

• Example. The set of multiples ofn

nZ = {m ∈ Z | m = nk for somek ∈ Z} = {nk | k ∈ Z}

•Definition 2.1.1 A setI of real numbers is called aninterval if I , ∅,I contains more than one element, and for everyx, y ∈ I such thatx < y,if za real number such thatx < z< y, thenz ∈ I .

• Closed bounded interval

[a,b] = {x ∈ R|a ≤ x ≤ b}

• Open bounded interval

(a,b) = {x ∈ R|a < x < b}

• Half-open, half-closed bounded intervals

(a,b] = {x ∈ R|a < x ≤ b}, [a,b) = {x ∈ R|a ≤ x < b}

• Unbounded intervals

(−∞,b] = {x ∈ R|x ≤ b}, (−∞,b) = {x ∈ R|a ≤ x < b},(a,∞) = {x ∈ R|a < x}, [a,∞) = {x ∈ R|a ≤ x},

• Example

• Sets containing finitely many elements are calledfinite sets.

• The number of elements of a finite set is called thecardinality of the set.

• Notation. |A|• Sets containing infinitely many elements are calledinfinite sets.

• Example

2.1. LECTURE 5. SETS AND SUBSETS 21

2.1.2 Subsets

• Example

Definition 2.1.2 Let A and B be sets. A is a subsetof B if everyelement ofA is also an element ofB. If the setB has an element thatis not an element of its subsetA then the subsetA is called apropersubset.

• Notation.

A ⊆ B, (A is a subset ofB).

A ⊂ B (A is a proper subset ofB)

• Example

• Let U be an universal set. ThenA ⊆ B is the statement

“∀x ∈ U, if x ∈ A, thenx ∈ B”.

• Example

• Proposition 2.1.1 Let A, B andC be sets. IfA ⊆ B and B ⊆ C, thenA ⊆ C.

Proof:

• Example

• Definition 2.1.3 Two setsA and B are equal if they have the sameelements.

• Notation. A = B.

• Two setsA andB are equal if and only if every element ofA is an elementof B and every element ofB is an element ofA

(A = B)⇔ (A ⊆ ∧B ⊆ A)

• The statementA ⊆ B is false if there is an element ofA such that it is not anelement ofB.

• Example


22 CHAPTER 2. SETS

2.1.3 Complements

•Definition 2.1.4 Let A and B be sets. Thecomplement ofA in B isthe set

B− A = {x ∈ B|x < A}.• Example

• The setA does not need to be a subset of the setB.

• Let U be a universal set. The the complement ofA in U is simply called thecomplementof A.

• Notation.

A = U − A.

• The set that contains no elements is called theempty set.

• Notation. ∅• If U is an universal set, then

∅ = U andU = ∅.

•Theorem 2.1.1 LetU be a universal set andA andB be sets containedin U. Then

A ⊆ B iff B ⊆ A.

Proof:

• Example

2.1.4 Homework


• Exercises: 2.1[1,10,14]

2.2. LECTURE 6. COMBINING SETS 23

2.2 LECTURE 6. Combining Sets

2.2.1 Unions and Intersections

•Definition 2.2.1 Theunion of a setA and a setB is the set

A∪ B = {x|x ∈ A∨ x ∈ B}.

•Definition 2.2.2 Theintersectionof a setA and a setB is the set

A∩ B = {x|x ∈ A∧ x ∈ B}.• Example

• Two sets aredisjoint if they do not have common elemets, that is theirintersection is the empty set,

A∩ B = ∅.

• Venn diagrams

•

Theorem 2.2.1 Let A, B andC are sets. Then

1. A∪ B = B∪ A,

2. A∩ B = B∩ A,

3. (A∪ B) ∪C = A∪ (B∪C),

4. (A∩ B) ∩C = A∩ (B∩C),

5. A ⊆ (A∪ B),

6. (A∩ B) ⊆ A,

7. ∅ ⊆ A,

8. A∪ ∅ = A,

9. A∩ ∅ = ∅.Proof:

• Proposition. Let A, B ⊆ U. Then


24 CHAPTER 2. SETS

1. A− B = A∩ B.

2. A ⊆ B⇔ A∪ B = B

Proof:

•Theorem 2.2.2 Let A, B andC be sets. Then

1. A∩ (B∪C) = (A∩ B) ∪ (A∩C),

2. A∪ (B∩C) = (A∪ B) ∩ (A∪C).

Proof:

2.2.2 De Morgan’s Laws

• Example

Theorem 2.2.3 LetU be a universal set andA, B ⊆ U. Then

1. A∪ B = A∩ B,

2. A∩ B = A∪ B.

Proof:

2.2.3 Cartesian Products

•Definition 2.2.3 TheCartesian product of a setA and a setB is theset

A× B = {(a,b)|a ∈ A,b ∈ B}.• The Cartesian product is the set ofordered pairs (a,b), where the first

element is from the setA and the second element is fromB.

• Example

2.2.4 Homework


• Exercises: 2.2[22,24]


2.3. LECTURE 7. COLLECTION OF SETS 25

2.3 LECTURE 7. Collection of Sets

2.3.1 Power Set

• Example

• Definition 2.3.1 Thepower setof a setA is the set of all subsets ofA.

• Notation.

P(A) = {X|X ⊆ A}.

• Note that

∅ ∈ P(A) andA ∈ P(A).

• If the setA is finite, then the cardinality of the power set is

|P(A)| = 2|A|

• Example

2.3.2 Indexing Sets

• Example

• The indexing setis the set of subscripts that are used to distinguish the setsin a collection of sets.

• Notation.S = {Ai |i ∈ I } = {Ai}i∈I ,

whereI = {1,2, . . . , n}.

• Union of sets in a collection of sets

n⋃

i=1

Ai = {x ∈ U | ∃i ∈ I 3 x ∈ Ai}


26 CHAPTER 2. SETS

• Intersection of sets in a collection of setsn⋂

i=1

Ai = {x ∈ U | x ∈ Ai ∀i ∈ I }

• Example

• Infinite collection of sets.

• Union of sets in a collection of sets∞⋃

i=1

Ai = {x ∈ U | ∃i ∈ N 3 x ∈ Ai}

• Intersection of sets in a collection of sets∞⋂

i=1

Ai = {x ∈ U | x ∈ Ai ∀i ∈ N}

• Example

• The collection of setsS = {Ai}i∈I is increasing(or anascending chain) if

A1 ⊆ A2 ⊆ A3 ⊆ · · ·

• The collection of setsS = {Ai}i∈I is decreasing(or andescending chain) if

· · · ⊆ A3 ⊆ A2 ⊆ A1.

• Example

•

Definition 2.3.2 Let I be an indexing set andS = {Ai}i∈I be a collec-tion of sets.

1. Theunion of the collection is⋃

i∈IAi = {a| a ∈ A for somei ∈ I }

2. Theintersectionof the collection is⋃

i∈IAi = {a| a ∈ A for all i ∈ I }



• Notation. Let S be a collection of sets. Then the union and the intersectionare

∪A∈SA and∩A∈SA.

2.3.3 Partitions

•

Definition 2.3.3 A partition of a setA is a subsetP of the power setP(A) such that

1. if X ∈ P, thenX, ∅,2.

⋃X∈P X = A,

3. if X,Y ∈ P andX , Y, thenX ∩ Y = ∅.That is, a partition is a collection of non-empty disjoint subsets ofA thatcover the setA.

• Remark. A partition of a set divides the set into different disjoint nonemptysubsets so that every element of the set is in one of the subsets and noelement is in more than one.

• Example

2.3.4 The Pigeonhole Principle

•Theorem 2.3.1 Let A andB be finite disjoint sets. Then

|A∪ B| = |A| + |B|Proof: Rigorous proof later.

•

Corollary 2.3.1 Let S = {Ai}ni=1 be a collection of finite mutually dis-joint sets. Then ∣∣∣∣∣∣∣

n⋃

i=1

Ai

∣∣∣∣∣∣∣ =

n∑

i=1

|Ai |


28 CHAPTER 2. SETS

•Corollary 2.3.2 Let A andB be finite sets. Then

|A∪ B| = |A| + |B| − |A∩ B|• Pigeonhole Principle. If n objects are placed ink containers andn > k,

then at least one container will have more than one object in it.

• Let S = {Ai}ni=1 be a collection of finite mutually disjoint sets andA =⋃ni=1 Ai.

If |A| = k andk > n, then, for somei, |Ai | ≥ 2.

• Example

•Theorem 2.3.2 LetS = {Ai}ni=1 be a collection of finite mutually disjointsets andA =

⋃ni=1 Ai. If |A| > nr for some positive integerr, then, for

somei, |Ai | ≥ r + 1.

Proof: Exercise.

• Example

2.3.5 Cantor Set

• Let A = [0,1].

• Let A1 = [0, 13] ∪ [ 2

3,1]

• Let A2 = [0, 19] ∪ [ 2

9,39] ∪ [ 6

9,79] ∪ [ 8

9,1]

• DefineAn by removing the middle third from each of the closed intervalsthat make the setAn−1.

• EachAn is the union of 2n closed intervals

• Each interval that makesAn has length13n

• The total length of all intervals that makeAn is(

23

)n

• The collection of sets{An}∞n=1 forms a descending chain

An+1 ⊂ An, ∀n ∈ N



• TheCantor set is defined by

C =

∞⋂

n=1

An

• The endpoints of all intervals in each of the setsAn are inC

0,1,13,23,19,29,79,89, · · · ∈ C

• Cantor set is infinite.

• Cantor set is bounded.

• The “size” (length) of the Cantor set is zero. Since the length ofC is smallerthan the length ofAn for anyn ∈ N, which is (23)n. Thus it is smaller than(2

3)n for anyn, and is, therefore, zero.

• Cantor set is uncountable. So, it is “larger” than the set of integers.

2.3.6 Homework


• Exercises: 2.3[16,26,28]


30 CHAPTER 2. SETS


Chapter 3

Functions

3.1 LECTURE 8. Definition and Basic Properties

•Definition 3.1.1 Let A andB be nonempty sets. Afunction f from Ato B is a rule that assigns to each element of the setA one and only oneelement of the setB.The setA is thedomain of f and the setB thecodomain.

• Notation.

f : A→ B

and

for eacha ∈ A, f (a) = b or af7→ b.

• Example

• The identity function on A is the functioniA : A→ A defined by

iA(x) = x for anyx ∈ A.

It is also denoted by IdA or idA.

31

32 CHAPTER 3. FUNCTIONS

3.1.1 Image of a Function

•Definition 3.1.2 Let f : A→ B be a function andX ⊆ A. Theimageof the setX under the functionf is the set

f (X) = {y ∈ B|y = f (x) for somex ∈ X}• Diagram.

• Remark.

f (∅) = ∅ for any functionf .

•Definition 3.1.3 Let f : A → B be a function. Theimage (or therange) of the functionf is the set

Im( f ) = f (A) = {y ∈ B|y = f (x) for somex ∈ A}• Remark.

Im( f ) ⊆ B for any functionf .

• Diagram.

•Definition 3.1.4 Let f : A → B be a function. Thegraph of thefunction f is the set

Γ( f ) = {(a,b) ∈ A× B|b = f (a),a ∈ A}• Diagram.

• Remark.

Γ( f ) ⊆ A× B.

• Remark. The functionsf : A→ B andg : A→ f (A) such thatg(x) = f (x)for anyx ∈ A aredifferentif f (A) , B.

•Definition 3.1.5 Two functionsf and g are equal if they have thesame domain and the same codomain and iff (x) = g(x) for anyx in thedomain.


3.1. LECTURE 8. DEFINITION AND BASIC PROPERTIES 33

• Examples

• Sine function

sin :R→ R, xsin7→ sin(x)

• Greatest integer function.

f : R→ Z, f (x) = [x] is the greatest integer≤ x

• Finding the image of a function. If f : A→ B is a function andy ∈ B, theny ∈ Im( f ) if and only if there exists an elementx ∈ A such thatf (x) = y.

• Example.

•Theorem 3.1.1 Intermediate Value Theorem.Let A, B ⊆ R be sub-sets of real numbers andf : A→ B be a function. Let[a,b] ⊆ A and fbe continuous on the interval[a,b]. If y is a number betweenf (a) andf (b), then there is a real numberx ∈ [a,b] such thatf (x) = y.

Without proof.

• Graphical illustration.

• Example

•

Proposition 3.1.1 LetA andB be sets andX andY be subsets ofA suchthat

X ⊆ Y ⊆ A.

Let f : A→ B be a function. Then

f (X) ⊆ f (Y).

Proof: Easy.

• The converse of this proposition is false.

• Example.



•

Proposition 3.1.2 LetA andB be sets andX andY be subsets ofA. Letf : A→ B be a function. Then

1. f (X ∪ Y) = f (X) ∪ f (Y).

2. f (X ∩ Y) ⊆ f (X) ∩ f (Y).

Proof:

• Example.

3.1.2 Inverse Image

•

Definition 3.1.6 Let A and B be sets andW be a subset ofB. Letf : A → B be a function. The theinverse image of the setW withrespect to f is the set

f −1(W) = {x ∈ A| f (x) ∈W}• Diagram.

• The inverse imagef −1(W) is the set of all elements of the domain off thatare mapped to elements ofW.

• Remarks. Inverse image is a subset of the domain.

f −1(W) ⊆ A

• W is not necessarily a subset of the image Im(f ) of the functionf .

• The symbolf −1 doesnot refer to the inverse function, which might not evenexist.

• Example

•

Proposition 3.1.3 Let A and B be sets andW and Z be subsets ofB.Let f : A→ B be a function. Then

1. f −1(W∪ Z) = f −1(W) ∪ f −1(Z)

2. f −1(W∩ Z) = f −1(W) ∩ f −1(Z)

Proof:


3.1. LECTURE 8. DEFINITION AND BASIC PROPERTIES 35

•Proposition 3.1.4 Let A andB be sets andf : A→ B be a function. IfA is a finite set, thenIm( f ) is a finite set and

|Im( f )| ≤ |A|.Proof: Easy.

• Example

3.1.3 Homework


• Exercises: 3.1[2,4,11]



3.2 LECTURE 9. Surjective and Injective Func-tions.

3.2.1 Surjective Functions

•Definition 3.2.1 Let f : A → B be a function. Thenf is surjective(or a surjection) if the image off is equal to the codomain off , that is

Im( f ) = B.

• To prove that a function is surjective one needs to prove that for any elementy in the codomainB there is an element in the domainx ∈ A such thatf (x) = y.

• Diagram.

• Surjection is also called anonto mapping.

• Example

• Let A andB be sets. The functionπ : A× B→ A defined by

π(a,b) = a

is aprojection of A× B ontoA.

• To prove that the functionf : A→ B is not surjective one needs to show thatthere exists an elementy ∈ B in the codomainB such that for any elementx ∈ A in the domain,f (x) , y.

• Examples.

3.2.2 Injective Functions

• Example

•Definition 3.2.2 Let f : A→ B be a function. Thenf is injective (oran injection) if

for anya1,a2 ∈ A, if a1 , a2, then f (a1) , f (a2).


3.2. LECTURE 9. SURJECTIVE AND INJECTIVE FUNCTIONS. 37

• An injection takes different elements of the domainA to different elementsof the codomainB.

• Injection is also called anone-to-onemapping.

• Diagram.

• Example

• To prove that a functionf : A→ B is injectiveone needs to prove that

for all a1,a2 ∈ A, if f (a1) = f (a2), thena1 = a2.

• Examples.

• To prove that a functionf : A→ B is not injectiveone needs to show that

there exista1,a2 ∈ A such thata1 , a2 and f (a1) = f (a2).

(In other words,f maps two different elements ofA to the same element ofB.)

• Example.

3.2.3 Bijective Functions

• Definition 3.2.3 A function that is both injective and surjective isbi-jective (or a bijection).

• Examples.

• Definition 3.2.4 Let A be a set. A bijectionf : A→ A is a permuta-tion of A.

• Example

3.2.4 Homework


• Exercises: 3.2[2,13,17]



3.3 LECTURE 10. Composition and Inverse Func-tions

•

Definition 3.3.1 Let A and B be nonempty sets. The sets of all func-tions fromA to B is denoted by

F(A, B) = { f | f : A→ B}.

If A = B, then we denote

F(A,A) = F(A).

3.3.1 Composition of Functions

•

Definition 3.3.2 LetA, B, andC be nonempty sets, and letf ∈ F(A, B)and g ∈ F(B,C). Thecomposition of f and g is the functiong ◦ f ∈F(A,C) defined by

(g ◦ f )(x) = g( f (x)),∀x ∈ A

• Notation. The compositiong ◦ f is also denoted simply byg f .

• If A = B = C, then bothg ◦ f and f ◦ g are defined.

• Examples.

•Proposition 3.3.1 Let f : A→ B be a function. Then

f ◦ iA = iB ◦ f = f .

Proof: Exercise.

•

Proposition 3.3.2 Let f ∈ F(A, B) andg ∈ F(B,C). Then:

1. If f andg are surjections, theng ◦ f is a surjection.

2. If f andg are injections, theng ◦ f is an injection.

3. If f andg are bijections, theng ◦ f is a bijection.

Proof:


3.3. LECTURE 10. COMPOSITION AND INVERSE FUNCTIONS 39

• Corollary 3.3.1 Let A be a nonempty set. Letf andg be permutationsof A. Theng ◦ f is a permutation ofA.

Proof: Exercise.

•Proposition 3.3.3 LetA, B, C andD be nonempty sets. Letf ∈ F(A, B),g ∈ F(B,C), h ∈ F(C,D). Then

(h ◦ g) ◦ f = h ◦ (g ◦ f ).

Proof:

• Notation. One can write without ambiguity simply

h ◦ g ◦ f = (h ◦ g) ◦ f = h ◦ (g ◦ f ).

• If we haven functions f1, f2, . . . , fn, then one can writef1 ◦ f2 ◦ · · · ◦ fn fortheir composition.

3.3.2 Inverse Functions

•

Definition 3.3.3 Let A andB be nonempty sets andf ∈ F(A, B). Thenf is invertible if there is a functionf −1 ∈ F(B,A) such that

f ◦ f −1 = iB

andf −1 ◦ f = iA.

If f −1 exists, it is called theinverseof f .

• If f is invertible, thenf −1 is invertible and

( f −1)−1 = f

• Proposition 3.3.4 There is only one function that can be the inverse ofa function f .

Proof:

• Let A be a set andiA : A→ A be the identity function. TheniA is invertibleand

i−1A = iA.



• Examples.

• Theorem 3.3.1 Let A and B be sets, and letf ∈ F(A, B). Then f isinvertible if and only iff is a bijection.

Proof:

• Examples.

3.3.3 Homework


• Exercises: 3.3[1,20]


Chapter 4

Binary Operations and Relations

4.1 LECTURE 11. Binary Operations

•• Example

•Definition 4.1.1 Let A be a nonempty set. Abinary operation is afunction

f : A× A→ A.

• Notation.

f (a,b), a · b, a? b, a + b, a ∗ b, a ◦ b, · · ·

• The elementf (a,b) must be in the setA.

• Examples.

4.1.1 Associate and Commutative Laws

•Definition 4.1.2 A binary operation∗ on A is associativeif

(a ∗ b) ∗ c = a ∗ (b ∗ c), ∀a,b, c ∈ A

•Definition 4.1.3 A binary operation∗ on A is commutative if

a ∗ b = b ∗ a, ∀a,b ∈ A

41

42 CHAPTER 4. BINARY OPERATIONS AND RELATIONS

• To prove that an operation is associative or commutative one needs to provethe property forall elements ofA.

• To prove that an operation isnot associative or commutative one needs toshow that the property does not hold forsomeelements ofA.

• Examples.

• Union of sets onP(U)

• Intersection of sets onP(U)

• Composition of functions onF(A)

• Multiplication and addition onF(R)

• Matrix multiplication onM2(R)

• Matrix addition onM2(R)

4.1.2 Identities

•

Definition 4.1.4 Let ∗ be a binary operation on the setA. An elemente is an identity elementof A with respect to∗ if

a ∗ e = e∗ a = a

for anya ∈ A.

• Not all sets have identity elements with respect to a given binary operation.

• Examples.

• Proposition 4.1.1 If e is an identity element of the setA with respect toa binary operation∗ on A, thene is unique.

Proof: By contradiction.


4.1. LECTURE 11. BINARY OPERATIONS 43

4.1.3 Inverses

•

Definition 4.1.5 Let∗ be a binary operation on the setA with identitye. We say thata ∈ A is invertible with respect to∗ if there existsb ∈ Asuch that

a ∗ b = b ∗ a = e.

If

a ∗ b = b ∗ a = e,

thenb is an inverseof a with respect to∗.

• Examples.

•Proposition 4.1.2 Let ∗ be an associative binary operation on a setAwith an identity elemente. If a ∈ A has an inverse with respect to∗,then that inverse is unique.

Proof: Exercise.

• Notation. The inverse ofa is denoted bya−1.

4.1.4 Closure

•Definition 4.1.6 LetA be a nonempty set andX ⊆ A. Let∗ be a binaryoperation on the setA. If for any x, y ∈ X, we havex ∗ y ∈ X, thenX isclosed inA under ∗.

• If X ⊆ A is closed inA under∗, then∗ is also a binary operation onX.

• Examples.

• To prove thatX ⊆ A is not closed under a binary operation∗, we need toprove that there existx, y ∈ X such thatx ∗ y < X.

• Examples.


4.1.5 Groups

•

Definition 4.1.7 A set with a binary operation is a group if

1. the binary operation is associative,

2. there is an identity element,

3. every element has an inverse.

• Examples.

4.1.6 Homework


• Exercises: 4.1[2,5,19,22,36,37]


4.2. LECTURE 12. EQUIVALENCE RELATIONS 45

4.2 LECTURE 12. Equivalence Relations

4.2.1 Relations

• Definition 4.2.1 A relation Ron a setA is a subset ofA× A.

• Notation. If (a,b) ∈ R, then we writeaRb

• Examples.

4.2.2 Properties of Relations

• The relation

R = {(a,a)|a ∈ A}

is called thediagonalof A× A.

•

Definition 4.2.2 LetR be a relation on a setA. Then

1. R is reflexive if ∀a ∈ A,

aRa,

2. R is symmetric if ∀a,b ∈ A,

if aRb, thenbRa,

3. R is transitive if ∀a,b, c ∈ A,

if aRbandbRc, thenaRc,

4. R is antisymmetric if ∀a,b ∈ A,

if aRbandbRa, thena = b.

• Examples.



4.2.3 Equivalence Relations

•

Definition 4.2.3 A relationR on a setA is anequivalence relationifit is

1. reflexive,

2. symmetric and

3. transitive.

• Notation. An equivalence relation is denoted by

a ∼ b

read as

“a is equivalent tob”

• Examples.

• Congruence modn is a relationR onZ+ defined by

aRbif a− b = nk for somek ∈ Z

written

a ≡ b(modn)

• Example

4.2.4 Equivalence Classes

•

Definition 4.2.4 Let∼ be an equivalence relation on a setA. Then foranya ∈ A the set

[a] = {x ∈ A | x ∼ a}is theequivalence classof a.

• Equivalence class ofa is the set of all elements equivalent toa.



• All elements of an equivalence class are equivalent to each other.

• Examples.

• An equivalence relation divides the set intodisjoint equivalence classes.

• Theorem 4.2.1 The set of equivalence classes of an equivalence rela-tion ∼ on a nonempty setA forms a partition ofA.

Proof:

1. Equivalence classes are subsets ofA.

2. Equivalence classes are nonempty.

3. Every element ofA is in some equivalence class.

4. Different equivalence classes are disjoint. In other words,

if [ a] , [b], then [a] ∩ [b] = ∅,or

if [ a] ∩ [b] , ∅, thena ∼ b and [a] = [b].

�

•

Theorem 4.2.2 LetP be a partition of a nonempty setA. Let R be arelation onA defined by

aRbif a andb are in the same element of the partition.

ThenR is an equivalence relation.

Proof:

1. R is reflexive.

2. R is symmetric.

3. R is transitive.

�

• There is a bijection between the set of all equivalence relations ofA and theset of all partitions ofA.


4.2.5 Partial and Linear Ordering

• Example. Orderingof the real numbers.

aRbif a ≤ b

R is reflexive, transitive andantisymmetric.

•

Definition 4.2.5 A relationR on a setA is a partial ordering on A ifR is

1. reflexive,

2. transitive and

3. antisymmetric.

A set with a partial ordering is apartially ordered set.

• The relation< is nota partial ordering onR.

• Let A be a set andP(A) be the power set ofA. Let R be a relation onP(A)defined by

XRYif X ⊆ Y.

ThenR is a partial ordering onP(A).

• Example. Linear orderingof real numbers.

∀x, y ∈ R, eitherx ≤ y or y ≤ x.

•

Definition 4.2.6 LetA be a set andRbe a partial ordering onA. ThenR is a linear ordering on A if

∀a,b ∈ A, eitheraRbor bRa.

A set with a linear ordering is alinearly ordered set.

• Examples.


4.2.6 Homework


• Exercises: 4.2[10,14,15,17]


Chapter 5

The Integers

5.1 LECTURE 13. Axioms and Basic Properties

5.1.1 The Axioms of the Integers

•

Definition 5.1.1 The setZ of integers is a set with two binary opera-tions: addition, +, andmultiplication , ·, with the properties

A1 Associativity of addition:

(x + y) + z = x + (y + z), ∀x, y, z ∈ Z

A2 Commutativity of addition :

x + y = y + x, ∀x, y ∈ Z

A3 Additive identity :

∃ 0 ∈ Z such that x + 0 = x, ∀x ∈ Z

A4 Additive inverse:

∀x ∈ Z, ∃(−x) ∈ Z such that x + (−x) = 0

51

52 CHAPTER 5. THE INTEGERS

Definition 5.1.2

A5 Associativity of multiplication :

(x · y) · z = x · (y · z), ∀x, y, z ∈ Z

A6 Commutativity of multiplication :

x · y = y · x, ∀x, y ∈ Z

A7 Multiplicative identity :

∃ 1 ∈ Z such that x · 1 = x, ∀x ∈ Z, and 1 , 0

A8 Distributivity :

x · (y + z) = x · y + x · z, ∀x, y, z ∈ Z• Notation.

x + (−y) = x− y, x · y = xy

• These properties of integers areaxioms.

• The multiplicative inverses of integers do not exist, in general.

• The axiom 1, 0 is needed for a nontrivial theory.

• If 1 = 0, thenx = 0 for anyx ∈ Z. Therefore,Z = {0} consists of only oneelement.

•

Proposition 5.1.1 Leta,b, c ∈ Z. Then

P1 If a + b = a + c, thenb = c

P2 a0 = 0

P3 (−a)b = a(−b) = −(ab)

P4 −(−a) = a

Proof:


5.1. LECTURE 13. AXIOMS AND BASIC PROPERTIES 53

•

Proposition 5.1.2 Leta,b, c ∈ Z. Then

P5 (−a)(−b) = ab

P6 a(b− c) = ab− bc

P7 (−1)a = −a

P8 (−1)(−1) = 1

Proof:

•Definition 5.1.3 The set of positive integersZ+ is defined by

Z+ = {n ∈ Z | n = 1 + · · · + 1︸︷︷︸n

}

• Notation.

Z+ = N

•

Definition 5.1.4 The setZ+ has the properties

A9 Closure. Z+ is closed with respect to addition and multiplication:

if x, y ∈ Z+, thenx + y ∈ Z+ andxy ∈ Z+

A10 Trichotomy Law . For every integerx ∈ Z exactly one of thefollowing statements is true:

1. x ∈ Z+,

2. −x ∈ Z+,

3. x = 0.

• Proposition 5.1.3 If x ∈ Z andx , 0, thenx2 ∈ Z+.

Proof: Follows from closure ofZ+.

5.1.2 Inequalities

• Definition 5.1.5 Let x, y ∈ Z. Thenx < y if y− x ∈ Z+.

Notation.

x < y is read “x is less thany”

if x < y, theny > x

if x < y or x = y, thenx ≤ y

if x > y or x = y, thenx ≥ y

if x ≤ y, theny ≥ x

• If x ∈ Z+, thenx > 0.

•

Proposition 5.1.4 Leta,b, c ∈ Z.

Q1 Exactly one of the following holds:

a < b, b < a or b = a.

Q2 If a > 0, then−a < 0. (If a < 0, then−a > 0.)

Q3 If a > 0 andb > 0, thena + b > 0 andab> 0.

Q4 If a > 0 andb < 0, thenab< 0.

Q5 If a < 0 andb < 0, thenab> 0.

Q6 If a 0, thenac< bc.

Q9 If a bc.

Proof: Exercise.

5.1.3 The Well-Ordering Principle

•

Definition 5.1.6 A11. The Well-Ordering Principle. Everynonempty subset ofZ+ has a smallest element. That is, ifS is anonempty subset ofZ+, then

there isa ∈ S such thata ≤ x for all x ∈ S.

5.1. LECTURE 13. AXIOMS AND BASIC PROPERTIES 55

• Proposition 5.1.5 There is no integerx such that0 < x < 1.

Proof:

1. By contradiction.

2. Suppose thata ∈ Z+ is the smallest integer such that 0< a < 1.

3. Thena2 < a (contradiction).

• Corollary 5.1.1 The number1 is the smallest element ofZ+.

Proof: Exercise.

• Corollary 5.1.2 The only integers having multiplicative inverses inZare1 and−1.


5.1.4 Homework


5.2 LECTURE 14. Induction

5.2.1 Induction: A Method of Proof

• Induction is a method for proving statements about the positive integers.

• Let P(x) be an open sentence. The purpose of an induction proof is to showthat the statementP(n) is true for every positive integern ∈ Z+.

• Idea:

1. Verify (prove) thatP(1) is true.

2. Given a positive integerk ∈ Z+ for which P(k) is true, prove thatP(k + 1) is true.

3. This establishes thatP(n) is true for anyn ∈ Z+.

•

Theorem 5.2.1 First Principle of Mathematical Induction. LetP(n) be a statement about the positive integern. Suppose that

1. P(1) is true, and

2. If P(k) is true for ak ∈ Z+, thenP(k + 1) is true.

ThenP(n) is true for anyn ∈ Z+.

Proof: By contradiction and the Well-Ordering Principle.

• The assumption thatP(k) is true is theinduction hypothesis.

• Examples.

5.2.2 Other Forms of Induction

•

Theorem 5.2.2 Modified Form of First Principle of MathematicalInduction Let P(n) be a statement about the integern. Suppose thatthere is an integern0 ∈ Z such that

1. P(n0) is true, and

2. If P(k) is true for an integerk ≥ n0 , thenP(k + 1) is true.

ThenP(n) is true for any integern ≥ n0.


5.2. LECTURE 14. INDUCTION 57

Proof: Exercise.

• Example

•

Theorem 5.2.3 Second Principle of Mathematical Induction. LetP(n) be a statement about the positive integern. Suppose that

1. P(1) is true, and

2. If, for a positive integerk ∈ Z+, P(i) is true for every positiveintegeri ≤ k, thenP(k + 1) is true.

ThenP(n) is true for any positive integern ∈ Z+.

Proof: Exercise.

• A function f : Z+ → Z+ is definedrecursively if the value f (k) of f at apositive integerk is defined by the valuesf (1), f (2), . . . , f (k−1) of f at thepreceding positive integers.

f (k) = F( f (1), . . . , f (k− 1))

This equation is arecursion, or arecursive relation.

• Example. Let f : Z+ → Z+ be defined byf (1) = 1, f (2) = 5, and

f (n + 1) = f (n) + 2 f (n− 1).

Then

f (n) = 2n + (−1)n

Proof: By induction.

• Restatement of the First Principle of Mathematical Induction in the lan-guage of the set theory.

•

Theorem 5.2.4 Let S ⊆ Z+ be a subset of positive integers. Supposethat

1. 1 ∈ S, and

2. If k ∈ S, thenk + 1 ∈ S.

ThenS = Z+.



Proof: Exercise.

• Examplen∑

k=1

k =n(n + 1)

2

•Theorem 5.2.5 Let A be a set and{Bi}ni=1 be a finite collection of sets.Then

A⋂

n⋃

i=1

Bi

=

n⋃

i=1

(A

⋂Bi

)


5.2.3 The Binomial Theorem

•

Definition 5.2.1 Letn be a nonnegative integer. Thefactorial n! of nis defined by

0! = 1,

and

n! = n(n− 1) · · · 2 · 1 if n > 0.

•

Definition 5.2.2 Let n and r be nonnegative integers such that0 ≤r ≤ n. Thebinomial coefficient

(nr

)is defined by

(nr

)=

n!r!(n− r)!

=n(n− 1) · · · (n− r + 1)

r!

• Notation. (nr

)= Cn

r

• Note that (00

)=

(nn

)=

(n0

)= 1

• Also (nk

)=

(n

n− k

)


5.2. LECTURE 14. INDUCTION 59

• The binomial coefficient(nr

)is the number of ways to chooser objects from

a collection ofn objects.

• The binomial coefficient(nr

)is the number of subsets ofr elements in a set

with n elements.

• Identity . Let n, k ∈ Z+ such that 1≤ k ≤ n. Then(n + 1

k

)=

(nk

)+

(n

k− 1

)

Proof: Exercise.

•Theorem 5.2.6 Leta,b ∈ Z andn ∈ Z+. Then

(a + b)n =

n∑

k=0

(nk

)an−kbk


• Example

•Corollary 5.2.1 Letn ∈ Z be a nonnegative integer,n ≥ 0. Then

n∑

k=0

(nk

)= 2n

Proof: Apply the binomial theorem to (1+ 1)n.

5.2.4 Homework


• Exercises: 5.2[1,2,4,26]



5.3 LECTURE 15. The Division Algorithm

5.3.1 Divisors and Greatest Common Divisors

•

Definition 5.3.1 Leta andb be integers.We sayb dividesa if there is an integerc such that

a = bc.

The integera is divisible byb andc.The integersb andc are factors of a.

• Notation.

b | a (read “b dividesa”)

• Example

•

Proposition 5.3.1 Leta,b, c ∈ Z.

1. If a|1, thena = 1 or a = −1.

2. If a|b andb|a, thena = b or a = −b.

3. If a|b anda|c, thena|(bx+ cy) for anyx, y ∈ Z.

4. If a|b andb|c, thena|c.

Proof: Exercise.

• Example

•

Definition 5.3.2 Leta andb be integers, not both zero.A common divisor of a andb is an integerc such thatc divides bothaandb.A greatest common divisorof a andb is a positive integerd such that

1. d is a common divisor ofa andb, and

2. for any integerc, if c is a common divisor ofa and b, thencdividesd.

• Notation. The greatest common divisor (gcd) of two integersa andb isdenoted by


5.3. LECTURE 15. THE DIVISION ALGORITHM 61

gcd(a,b) = (a,b)

• Note that for anya ∈ Z, a , 0,

gcd(a,0) = |a|

• Example

•Theorem 5.3.1 Division Algorithm. Let a be an integer andb be apositive integer. Then there exist unique integersq andr such that

a = bq+ r and 0 ≤ r < b.

Proof:

1. LetS = {a− bx|x ∈ Z} andS0 = {n ∈ S|n ≥ 0}.2. Show thatS0 , ∅. (If a ≥ 0, thena ∈ S0. If a < 0, thena− ba ∈ S0.)

3. Let r be the smallest element ofS0. Thenr ≥ 0.

4. Also,r = a− bq for someq ∈ Z.

5. Thenr < b, since ifr ≥ b, thenr − b < r is the smallest element ofS0

(contradiction).

6. Uniqueness. Assume∃ q1, r1 andq2, r2 and show thatq1 = q2 andr1 = r2.

�

•

Theorem 5.3.2 Leta andb be integers, not both zero. Then:

1. the greatest common divisord of a andb exists and is unique,

2. there exist integersx andy such that

d = ax+ by.

Proof:

1. LetS = {ax+ by|x, y ∈ Z} andS0 = {n ∈ S|n > 0}.2. Then±a, ±b ∈ S. So,S0 , ∅.3. Letd be the smallest element ofS0.

4. Thend ∈ S. Henced = ax+ by for somex, y ∈ Z.

5. Also, there existq, r ∈ Z such thata = dq+ r and 0≤ r < d.

6. Sincer = a(1− xq) + b(−yq), thenr ∈ S.

7. It is impossible thatr > 0 since thenr ∈ S0 andr < d (contradiction).Therefore,r = 0.

8. So,a = dq, or d|a. Similarly, d|b.

9. Letc ∈ Z be a common divisor ofa andb. Thena = cuandb = cv forsomeu, v ∈ Z.

10. Therefore,d = ax+ by = c(ux+ vy). So,c|d.

11. Thusd = gcd(a,b).

12. Uniqueness. Exercise.

�

5.3.2 Euclidean Algorithm

•

Lemma 5.3.1 Leta andb be integers, not both zero. If there exist twointegersq andr such that

a = bq+ r and 0 ≤ r < b,

then

gcd(a,b) = gcd(b, r)

Proof:

1. Letd = gcd(a,b).

2. Thend|r.3. For anyc ∈ Z, if c|b andc|r, thenc|a.

4. Sinced = ax+ by, thenc|d.

�

• Euclidean Algorithm is the following procedure for finding the gcd of twointegers.

5.3. LECTURE 15. THE DIVISION ALGORITHM 63

1. Let a and b be two positive integers anda > b. By repeating thedivision algorithm we get

a = bq1 + r1, where 0≤ r1 < b

b = r1q2 + r2, where 0≤ r2 < r1

r1 = r2q3 + r3, where 0≤ r3 < r2

2. Since each remainder is strictly less than the previous remainder, even-tually we get

rn−2 = rn−1qn + rn, where 0≤ rn < rn−1

rn−1 = rnqn+1 + rn+1, wherern+1 = 0

3. By Lemma we have

gcd(a,b) = gcd(b, r1) = gcd(r1, r2) = gcd(r2, r3) =

· · · = gcd(rn−1, rn) = gcd(rn,0) = rn

4. Thus, gcd(a,b) is the last nonzero remainder that occurs in this proce-dure.

5. If r1 = 0, then gcd(a,b) = b.

• Example.

5.3.3 Relatively Prime Integers

•Definition 5.3.3 Two integersa andb, not both zero, arerelativelyprime if

gcd(a,b) = 1

• Two integers are relatively prime if they have no common divisors exceptfor 1 and−1.

• Example


•Theorem 5.3.3 Two integersa andb are relatively prime if and only ifthere exist integersx andy such that

ax+ by = 1

Proof:

1. If gcd(a,b) = 1, thenax+ by = gcd(a,b) = 1 for somex, y ∈ Z.

2. Letd = gcd(a,b).

3. If ax+ by = 1 for somex, y ∈ Z, thend|(ax+ by).

4. So,d|1.

5. Thusd = 1.

�

• Theorem 5.3.4 Leta,b, c ∈ Z. If a|(bc) andgcd(a,b) = 1, thena|c.

Proof:

1. We havea|(bc).

2. So,az= bc for somez ∈ Z andax+ by = 1 for somex, y ∈ Z.

3. Thenc = c(ax+ by) = a(cx+ zy).

4. Therefore,a|c.

�

5.3.4 Homework




5.4. LECTURE 16. PRIMES AND UNIQUE FACTORIZATION 65

5.4 LECTURE 16. Primes and Unique Factoriza-tion

5.4.1 Prime Numbers

•Definition 5.4.1 A prime number is an integerp ∈ Z greater than1whose only divisors are±1 and±p.A compositenumber is an integerp ∈ Z greater than1 that is not prime(that is, it has divisors different from±1 or ±p).

• Remark. 1 is not a prime number.

• Example

•Lemma 5.4.1 Let n ∈ Z+ be an integer greater than1. Thenn iscomposite if and only if there exist integersa andb such that

n = ab, where1 < a < n and1 1 andn is not divisible by a prime number}.2. Claim:T = ∅.3. SupposeT , ∅.4. Then∃ smallestn0 ∈ T.

5. Thenn0 is not prime.

6. Thenn0 = abwhere 1< a,b < n0. Therefore,a|n0.

7. Sincea < n0, thena < T.

8. Hence, there is a primep such thatp|a.

9. Sincep|a anda|n0, thenp|n0.

10. Contradiction ton0 ∈ T.

11. ThusT = ∅.


�

•Proposition 5.4.2 Leta,b ∈ Z and p be a prime number. Then

if p|(ab), thenp|a or p|b.

Proof:

1. Supposep does not dividea.

2. Thena , 0 and gcd(a, p) = 1.

3. Therefore,p|b. (By the theorem that says that ifa|(bc) and gcd(a,b) =

1, thena|c.)

�

•

Corollary 5.4.1 Let m ∈ Z+ be a positive integer,a1, . . . , am ∈ Z beintegers,a =

∏ni=1 ai be the product of these integers, andp be a prime

number. Then

if p|a, thenp|ak for somek, 1 ≤ k ≤ m.

Proof: Exercise. Induction.

•Corollary 5.4.2 Letm ∈ Z+ be a positive integer,a ∈ Z be an integer,and p be a prime number. Then

if p|am, thenp|a .

Proof: Exercise.

• Example

5.4.2 Unique Factorization

•Theorem 5.4.1 Unique factorization Theorem.Let n ∈ Z be aninteger greater than1. Then eithern is a prime number, or it can bewritten as a product of prime numbers.The product is unique, except for the order in which the factors appear.

Proof:

1. Existence. By induction.


5.4. LECTURE 16. PRIMES AND UNIQUE FACTORIZATION 67

2. Letn ∈ Z andn > 1. Let P(n) be the statement:n is either prime or aproduct of primes.

3. P(2) is true.

4. Letk > 2. Suppose thatP(2), . . . P(k) are true.

5. Claim: P(k + 1) is true.

6. If k + 1 is prime, thenP(k + 1) is true.

7. If k + 1 is not prime, then there exista,b ∈ Z such thatk + 1 = aband1 < a < k + 1 and 1< b < k + 1.

8. Botha andb are either primes or product of primes.

9. Thus,k + 1 is a product of primes.

10. So,P(k + 1) is true.

11. Thus,P(n) is true∀n ∈ Z.

12. Uniqueness. By contradiction.

13. Supposen = p1 · · · ps = q1 · · · qr with some primesqi andpj.

14. Claim:s = r and after some renumberingpi = qi, ∀i ∈ {1, . . . , s}15. Supposes≤ r.

16. Sincep1|n, thenp1|(q1 · · · qr).

17. Therefore,p1|qj for somej. Renumber theqi ’s so thatq j = q1.

18. Sincep1 andq1 are primes thenp1 = q1.

19. Thenp2 · · · ps = q2 · · · qr .

20. Also,p2|(q2 · · · qr).

21. So,p2|qm for somem. Renumber theqi ’s so thatqm = q2.

22. Thusp2 = q2.

23. Repeating this process, afters steps, we getp1 = q1, . . . , ps = qs.

24. If s< r, then we have 1= qs+1 · · · qr (contradiction).

25. Thuss = r.

26. Therefore,pi = qi for all i.

�

• Standard Form. Let n ∈ Z be an integer greater than 1. By grouping therepeated primes we can write

n = pm11 · · · pmr

r

where the exponentsm1, . . . ,mr ∈ Z+ are positive integers and

p1 < p2 < · · · < pr

aredistinctprimes.

• Example

5.4.3 Euclid Theorem

• Theorem 5.4.2 Euclid Theorem. There are infinitely many primenumbers.


1. Let {p1, · · · pr} be the set of primes be andn = p1 · · · pr be their prod-uct.

2. Letm = n + 1. Thenm has a prime divisor.

3. Let the primepi with somei be such thatpi |m.

4. Also, we havepi |n.

5. Therefore,pi |(m− n). Sopi |1 (contradiction).

6. Thus, there are infinitely many primes.

�

5.4.4 Homework


• Exercises: 5.4[10,20]

5.5. LECTURE 17. CONGRUENCES 69

5.5 LECTURE 17. Congruences

5.5.1 Congruences and Their Properties

• Example. There are nox, y ∈ Z such thatx2 = 4y + 3.�

•Definition 5.5.1 Let n ∈ Z+ be a positive integer. Two integersa,b ∈Z are congruent modulon if the differencea − b is divisible byn (orn|(a− b))

• Notation.

a ≡ b(modn) read “a is congruent tob modulon”

• Congruence modn defines an equivalence relation onZ.

Proposition 5.5.1 Leta,b, c ∈ Z andn ∈ Z+. Then

1. a ≡ a(modn).

2. If a ≡ b(modn), thenb ≡ a(modn).

3. If a ≡ b(modn) andb ≡ c(modn), thena ≡ c(modn).

Proof: Exercise.�

•

Theorem 5.5.1 Leta,b, c,d ∈ Z andn ∈ Z+. Then

1. If a ≡ b(modn) andc ≡ d(modn), thena+ c ≡ b+ d(modn) andac≡ bd(modn).

2. If ab≡ ac(modn) andgcd(a,n) = 1, thenb ≡ c(modn)

Proof: Easy.�

• Corollary 5.5.1 Let a ∈ Z andn, k ∈ Z+. If a ≡ b(modn), thenak ≡bk(modn).

Proof: Exercise.�



•Proposition 5.5.2 Let a ∈ Z andn ∈ Z+, n > 1. Then there is exactlyone integerk such that

a ≡ k(modn) and0 ≤ k ≤ n− 1.

Proof: By Division Algorithm. Existence. Uniqueness.�

• Example. Show that for anyx ∈ Z, x2 ≡ 0(mod 4) orx2 ≡ 1(mod 4).

• Example. Show that there are nox, y ∈ Z such thatx2 = 4y + 3, or there isno x ∈ Z such thatx2 ≡ 3(mod 4).

•

Definition 5.5.2 Let a ∈ Z. Thecongruence classof a modn is theequivalence class[a] of a with respect to the equivalence relation ofcongruence modn:

[a] = {x ∈ Z | x ≡ a(modn)}

• For anya ∈ Z, a ≡ r(modn) for exactly one integerr such that 0≤ r ≤ n−1.

• Therefore,a ∈ [r].

• Thus there are exactlyn distinct congruence classes:

[0], [1], · · · , [n− 1].

• Note that

[n] = [0]

• To findr so thata ∈ [r], we simply compute the remainder whena is dividedby n, that isa = nq+ r.

• Example.

• There is more than one way to represent a congruence class.

• For congruences modn,

[a] = [b] if and only if a ≡ b(modn).


Proof: Exercise.�

• Remarks.

1. The congruence class [a] is a set, not an integer.

2. For an integera such hat 0≤ a < n, the congruence class [a] is the setof integers that give a remainder ofa when divided byn.

3. The equality of congruence classes [a] = [b] does not mean thata = b.It means thata ≡ b(modn) (or thata andb give the same remainderwhen divided byn).

5.5.2 The Set of Congruence Classes

•

Definition 5.5.3 Let n ∈ Z+, n > 1. Then the set of congruence classesis

Zn = {[0], [1], · · · , [n− 1]}The addition and multiplication onZn are defined by: for anya,b ∈ Z,if [a], [b] ∈ Zn, then

[a] + [b] = [a + b], [a][b] = [ab].

• The addition of congruence classes is well defined. That is it does not de-pend on the choice of the integers to represent the equivalence classes.


•

Theorem 5.5.2 Properties of Addition. Let n ∈ Z, n > 1, andZn bethe set of congruence classes. Leta,b, c ∈ Z. Then

1. Commutativity of addition:

[a] + [b] = [b] + [a]

2. Associativity of addition

([a] + [b]) + [c] = [a] + ([b] + [c])

3. Additive identity:

[a] + [0] = [a]

4. Additive inverse:

−[a] = [−a]

Proof: Exercise.�

• The multiplication of congruence classes is well-defined.



•

Theorem 5.5.3 Properties of Multiplication. Let n ∈ Z, n > 1, andZn be the set of congruence classes. Leta,b, c ∈ Z. Then

1. Commutativity of multiplication:

[a][b] = [b][a]

2. Associativity of multiplication

([a][b])[c] = [a]([b][c])

3. Multiplicative identity:

[a][1] = [a]

4. Distributive property:

[a]([b] + [c]) = [a][b] + [a][c]

Proof: Exercise.�

• Zn is a finite set.

• One can give the addition and multiplication tables.

• Example

•Theorem 5.5.4 Let n ∈ Z, n > 1, and let a ∈ Z. Then [a] has amultiplicative inverse if and only ifa andn are relatively prime, that isgcd(a,n) = 1.

Proof:

1. Suppose [a] has a multiplicative inverse.

2. Then there isx ∈ Z such that [a][ x] = [1].

3. Hence, [ax] = [1] andax = 1(modn).

4. Therefore,n|(ax− 1).

5. So,ax− 1 = nt for somet ∈ Z.



6. Therefore,ax+ n(−t) = 1.

7. So, by a previous theorem gcd(a,n) = 1.

8. Conversely, suppose gcd(a,n) = 1.

9. Then there existx, y ∈ Z such thatax+ ny = 1.

10. Hence [1]= [a][ x] + [n][y].

11. We have [n] = [0].

12. Thus [a][ x] = [1].

13. So, [x] is a multiplicative inverse of [a].

�

• To find the multiplicative inverse of [a] one has to solve the congruence

ax≡ 1(modn).

•

Theorem 5.5.5 Let n ∈ Z, n > 1, and leta,b ∈ Z. Let gcd(a,n) = 1.Then the congruence

ax≡ b(modn)

has a unique solution modn.

That is, there exists integerx ∈ Z such that

ax≡ b(modn)

and if there isy ∈ Z such that

ay≡ b(modn),

then

x ≡ y(modn).

Proof:

1. Since gcd(a,n) = 1, there existt, s ∈ Z such thatas + nt = 1, oras= 1(modn).

2. We getasb+ ntb = b, or a(sb) = b(modn)



3. Let x = sb.

4. Thenax≡ b(modn).

5. If y ∈ Z is such thatay≡ b(modn), thenax≡ ay(modn).

6. Sinceax≡ ay(modn) and gcd(a,n) = 1, thenx ≡ y(modn).

�

• If x ∈ Z is a solution to the congruenceax ≡ b(modn), then the set of allsolutions to the congruence is the congruence class [x].

• Example

5.5.3 Homework




Chapter 6

Infinite Sets

6.1 LECTURE 19. Countable Sets

6.1.1 Numerically Equivalent Sets

• Example

•Definition 6.1.1 Let A andB be subsets of a universal setU. The setsA andB arenumerically equivalent if there is a bijectionf : A→ B.

The setsA andB have the samecardinality .

• Notation. The numerical equivalence of the setsA andB is denoted by

A ≈ B

• Numerical equivalence is an equivalence relation on the power set ofU. LetA, B,C ⊆ U. Then

1. A ≈ A,

2. if A ≈ B, thenB ≈ A,

3. if A ≈ B andB ≈ C, thenA ≈ C.

77

78 CHAPTER 6. INFINITE SETS

•

Definition 6.1.2 Let A be a subset of a universal setU. ThenA isfinite if A = ∅ or there is a positive integern such thatA ≈ {1,2, . . . , n}.The integern is the number of elements ofA.

A set isinfinite if it is not finite, that is,A , ∅ and there is no positiveintegern such thatA ≈ {1,2, . . . , n}.

• Notation. The number of elements of a finite setA is denoted by

|A|

• Two finite setsA andBare numerically equivalent,A ≈ B, (or have the samecardinality) if and only if they have the same number of elements,|A| = |B|.

• The cardinality is also called thecardinal number.

• The cardinality of a finite set is the number of elements of the set.

• The cardinality of the empty set is

|∅| = 0

6.1.2 Countable Sets

• The set of positive integersZ+ is infinite.

•

Definition 6.1.3 A setA is countably infinite if it is numericallyequivalent toZ+, A ≈ Z+.

A set iscountable if it is either finite or countably infinite.

A set isuncountable if it is not countable.

• Alternatively, a set is countable if it is numerically equivalent to a subset ofZ+.

• Countably infinite sets have the same cardinality.

• The cardinality ofZ+ is denoted by

ℵ0 (read “aleph zero”)


6.1. LECTURE 19. COUNTABLE SETS 79

• The setZ is countably infinite.

• The set of even integers 2Z is countably infinite.

• Theorem 6.1.1 Let A be a countable set andB be a subset ofA. ThenB is countable.

Proof:

1. If A is finite, thenB is finite, and, hence, countable.

2. SupposeA is countably infinite.

3. If B is finite it is countable.

4. SupposeB is infinite.

5. There is a bijectionf : Z+ → A.

6. LetS = f −1(B) = {i ∈ Z+| f (i) ∈ B}7. Sincef is surjective, thenS is infinite.

8. LetS1 = S and letk1 be the smallest element ofS1.

9. LetS2 = S1 − {k1} andk2 be the smallest element ofS2.

10. Similarly, we define the sequence of subsetsSi+1 ⊆ Z+ by Si+1 =

Si − {ki}, whereki is the smallest element ofSi.

11. We havek1 < k2 < . . . .

12. ThenSi+1 = S1 − {k1, . . . , ki}.13. All Si are infinite.

14. ThenS = {k1, k2, . . . }.15. Leth : Z+ → S be defined byh(n) = kn .

16. Sincekn = km impliesn = m, thenh is injective.

17. Claim:h is surjective.

18. Lett ∈ S.

19. Letm be the number of integers inS that are less thant.

20. If m = 0, thent = k1.

21. If m, 0, then the integers less thant arek1, . . . , km.

22. Thent is the smallest element ofSm+1.

23. Therefore,t = km+1 = h(m).

24. Thus,h is surjective.

25. So,h is bijective.

26. Now, let us defineg = f ◦ h : Z+ → B by g(n) = f (h(n)) .

27. Since bothf andh are bijections, theng is a bijection.

28. Thus,B is countably infinite.

�

• Corollary 6.1.1 Every subset ofZ is countable.

• Theorem 6.1.2 Let f : Z+ → A be a surjection. ThenA is countabe.

Proof:

1. If A is finite, thenA is countable.

2. SupposeA is infinite.

3. LetS1 = f −1(A) .

4. ThenS1 = Z+.

5. Letn1 be the smallest element ofS1. Thenn1 = 1.

6. Leta1 = f (n1) andS2 = S1 − f −1({a1})

andn2 be the smallest element ofS2.

7. Similarly, we defineai = f (ni) and

Si = Si−1 − f −1({ai−1})andni be the smallest element ofSi.

8. ThenSi = Z+ − f −1({a1, . . . , ai−1}) .

9. We have1 = n1 < n2 < · · · < ni < ni+1 < · · ·

and we defineS = {n1,n2, . . . } .


10. Thenf |S : S→ A is a bijection.

11. Leth : Z+ → S be defined byh(i) = ni.

12. Sinceni = nj implies i = j, thenh is injective.

13. Claim:h is surjective.

14. Lett ∈ S.

15. Letmbe the number of integers inS that are less thant.

16. If m = 0, thent = n1.

17. If m, 0, then the integers less thant aren1, . . . , nm.

18. Thent is the smallest element ofSm+1.

19. Therefore,t = nm+1 = h(m).

20. Thus,h is surjective.

21. So,h is bijective.

22. Define a functiong = f ◦ h : Z+ → A by g(i) = f (h(i)).

23. Theng is bijective.

24. ThusA is countably infinite.

�

• Definition 6.1.4 Let U be a universal set. Asequenceof elements ofU is a functionf : Z+ → U.

• Notation. Let f : Z+ → U be a sequence. Letan = f (n). The sequence isdenoted by

(an)∞n=1 = (a1,a2, . . . )

• The elements of a sequence are not necessarily distinct (as in the sets).

• The elements of the sequence are ordered.

• A constant sequence is the sequence (a,a, . . . ).

• The image of the sequence,f (Z+), is countable.

• Let A be a countably infinite set and letf : Z+ → A be a bijection. Letan = f (n). Then the elements of the setAcan be identified with the sequence(a1,a2, . . . ) with no elements repeated, that isai , aj if and only if i , j.



• Conversely, if the elements of a setA can be written as a sequence, thenAis countable.

• Example

6.1.3 Unions of Countable Sets

• If S is a finite set of real numbers then the setZ ∪ S is countable.

•Theorem 6.1.3 Let {Ai}ni=1 be a finite collection of countable subsets

of a universal setU. Thenn⋃

i=1Ai is countable.

Proof:

1. There are bijectionsfi : Z+ → Ai.

2. LetA =n⋃

i=1Ai.

3. Define f : Z+ → A as follows:f (1) = f1(1), f (2) = f2(1), . . . f (n) = fn(1)f (n + 1) = f1(2), . . . f (2n) = fn(2)and so on.

4. For any j ∈ Z+ there are uniquem andr such thatj = mn+ r withm ∈ Z+ and 1≤ r ≤ n− 1.

5. Claim: f ( j) = f (mn+ r) = fr(m+ 1)

6. Claim: f is surjective.

7. Leta ∈ A.

8. Thena ∈ Ai for somei.

9. There isj ∈ Z+ such thatfi( j) = a.

10. Thenf (( j − 1)k + i) = fi( j) = a.

11. So,f is surjective.

12. So,A is countable.

�



6.1.4 The Rationals are Countable

• Theorem 6.1.4 The setQ of rational numbers is countably infinite.

Proof:

1. It is enough to restrict to the positive rationalsQ+.

2. Table of rationals.

3. Define f : Z+ → Q+ by diagonal argument.

4. Then f is a surjection.

5. So,Q is a countable set.

�

• Corollary 6.1.2 Any subset ofQ is countable.

Proof: Exercise.�

6.1.5 Cartesian Products of Countable Sets

• The setZ × Z is countable.

• More generally,

• Theorem 6.1.5 Let A and B be countable subsets of a universal setU. ThenA× B is countable.

Proof: Diagonal argument.�

•

Theorem 6.1.6 Let {Ai}ni=1 be a finite collection of countable subsetsof a universal setU. Then the set

n�

i=1

Ai = A1 × A2 × · · · × An

is countable.

Proof: Exercise.�



6.1.6 Homework


• Exercises: 6.1[3,7,15]


6.2. LECTURE 20. UNCOUNTABLE SETS 85

6.2 LECTURE 20. Uncountable Sets

• Any x ∈ R+ can be written in the decimal form

x = a0.a1a2 · · · =∞∑

k=0

10−kak ,

whereak ∈ {0,1,2,3,4,5,6,7,8,9} anda0 ∈ Z.

• Such an expansion is unique unless it ends in an infinite string of 9’s.

• Convention: Do not use decimal expansions ending in an infinite string of9’s.

• Binary expansion:

x = a0.a1a2 · · · =∞∑

k=0

2−kak ,

whereak ∈ {0,1} anda0 ∈ Z.

• Convention: Do not use binary expansions ending in an infinite string of1’s.

6.2.1 Uncountable Sets

• Theorem 6.2.1 The closed interval of real numbers[0,1] is uncount-able.

Proof:

1. By contradiction.

2. Diagonal argument.

�

• Corollary 6.2.1 The set of real numbersR is uncountable.

Proof: Since [0,1] ⊆ R.�

• For anya,b, c,d ∈ R

[a,b] ≈ [0,1], [c,d] ≈ [a,b], (a,b) ≈ (c,d), [a,b] ≈ (c,d)



• The cardinality ofR is called thepower of the continuum

|R| = c

• If A ≈ R, then|A| = c.

6.2.2 Cantor Theorem

•

Definition 6.2.1 Let A andB be subsets of a universal setU.We say thatA � B if there is an injectionf : A→ B.

We say thatA ≺ B if there is an injectionf : A→ B andA 0 B.

Let X andY be two sets andx andy be their cardinalities. Then

x ≺ y if X ≺ Y

and

x � y if X � Y.

•

Lemma 6.2.1 Let U be a universal set andA, B,C,D ⊆ U. Supposethat A ≈ B andC ≈ D. Then:

If A ≺ C, then B ≺ D.If A � C, then B � D.

Proof: Exercise.�

• Question: Are there any infinite sets with cardinality≺ |Z+| = ℵ0?

• Answer: No.

• Reason: Every infinite set contains a countably infinite subset.

• Question: Are there sets with infinite cardinalities different fromℵ0 andc?

• Answer: Yes.

• There are infinitely many different infinite cardinal numbers.


•Theorem 6.2.2 Cantor’s Theorem.Let A be a subset of a universalsetU andP(A) be the power set ofA. Then

A ≺ P(A).

Proof:

1. Claim: There is an injectionf : A→ P(A).

2. Leta ∈ A.

3. Let f (a) = {a}.4. Claim: Thenf is injective.

5. Claim: There is no surjectiong : A→ P(A).

6. Suppose such a surjection exists.

7. Let B = {a ∈ A|a < g(a)}.8. Sinceg is surjective, thenB ∈ g(A) = P(A).

9. So,∃a0 ∈ A such thatg(a0) = B.

10. Supposea0 ∈ B.

11. Thena0 < g(a0) = B (a contradiction).

12. Supposea0 < B. Thena0 ∈ g(a0) = B (a contradiction).

13. Thus, there is no surjection fromA to P(A).

14. ThusA ≺ P(A).

�

6.2.3 Continuum Hypothesis

• Question: Are there sets with cardinalityx such that

|Z+| = ℵ0 ≺ x ≺ c = |R|?

• Continuum Hypothesis. There is no setX with cardinalityx such thatℵ0 ≺ x ≺ c.

• Corollary . Let X ⊆ R. Then eitherX is finite, orX ≈ Z+, or X ≈ R.

• The continuum hypothesis is independent of the axioms of the set theoryand can be neither proved nor disproved from those axioms.


6.2.4 Schroeder-Bernstein Theorem

• Let U be a universal set. Then� is a relation onP(U).

• The relation� is reflexive and transitive.

• The relation� is neither symmetric nor antisymmetric. It is anti-symmetricon the set of equivalence classes [A].

• Theorem 6.2.3 Schroeder-Bernstein Theorem.LetA andB be sets.If A � B andB � A, thenA ≈ B.

Proof:

1. There are injectionsf : A→ B andg : B→ A.

2. Leta = a1 ∈ A.

3. If a = a1 ∈ g(B), then letb1 ∈ B such thata1 = g(b1).

4. If b1 ∈ f (A), then leta2 ∈ A such thatb1 = f (a2).

5. If a2 ∈ g(B), then letb2 ∈ B such thata2 = g(b2).

6. If b2 ∈ f (A), then leta3 ∈ A such thatb2 = f (a3).

7. Diagram.

8. For eacha ∈ A we construct an ordered listX(a) = (a1,b1,a2,b2, . . . )of ancestorsof a, where

ak = g(bk), bk = f (ak+1) .

9. ThenX(a) is either finite or infinite.

10. We define the subsets ofA by:

AA = {a ∈ A | X(a) is finite;∃ an oldest ancestory(a) ∈ A− g(B)}

AB = {a ∈ A|X(a) is finite;∃ an oldest ancestory(a) ∈ B− f (A)}

A∞ = {a ∈ A|X(a) is infinite; there is no oldest ancestor}

11. These subsets are mutually disjoint and define a partition ofA

A = AA ∪ A∞ ∪ AB .



12. Everya ∈ A must satisfy one of the three conditions.

13. Similarly, for each elementb ∈ B we define the list of its ancestorsX(b) = (b1,a2,b2,a3, . . . ) such that

ak = g(bk), bk = f (ak+1) .

14. This defines the partition ofB by subsetsBA, BB andB∞

B = BA ∪ B∞ ∪ BB .

15. Use the Lemma (proved below): The functions

f : AA→ BA, f : A∞ → B∞, g : BB→ AB, g : B∞ → A∞

are bijections.

16. Defineh : A→ B by

h(a) =

f (a), if a ∈ AA ∪ A∞

b, such thatg(b) = a, if a ∈ AB

17. Claim:h is a bijection.

18. Thus,A ≈ B.

�

• Lemma 6.2.2 The functionf : AA→ BA is a bijection.

Proof:

1. Sincef is an injection onA andAA ⊂ A, then f is an injection onAA.

2. We only need to prove thatf is a surjection onAA, that is, f (AA) = BA.

3. Leta ∈ AA andX(a) = (a,b1,a2,b2, . . . , y)

be the list of its ancestors. Theny ∈ A.

4. Letb = f (a) ∈ B. Then

X(b) = (b,a,b1,a2,b2, . . . , y) .



5. Therefore, the oldest ancestor ofb is in A. So,b = f (a) ∈ BA.

6. Thus,f (AA) ⊆ BA.

7. Letb ∈ BA andy be its oldest ancestor. Theny ∈ A. Thus, there is ana ∈ A such thatb = f (a).

8. ThenX(b) = (b,a,b1,a2,b2, . . . , y)

andX(a) = (a,b1,a2,b2, . . . , y) .

9. Thus, the oldest ancestor ofa is y ∈ A. That isa ∈ AA andBA ⊆ f (AA).

10. Thusf (AA) = BA, and f : AA → BA is a surjection, and, therefore, abijection.

�

• Lemma 6.2.3 The functiong : BB→ AB is a bijection.

Proof: The proof is the same as above.�

• Lemma 6.2.4 The functionsf : A∞ → B∞ and g : B∞ → A∞ arebijections.

Proof:

1. Leta ∈ A andX(a) = (a,b1,a2,b2, . . . )

be the list of its ancestors. Letb = f (a) ∈ B. Then

X(b) = (b,a,b1,a2,b2, . . . ) .

2. If a ∈ A∞, then the listX(a) is infinite. Therefore, the listX(b) isinfinite.

3. Thus,b = f (a) ∈ B∞, that is f (A∞) ⊆ B∞.

4. If b ∈ B∞, then the list

X(b) = (b,a,b1,a2,b2, . . . )

is infinite, and there isa ∈ A such thatb = f (a).


5. Then the listX(a) = (a,b1,a2,b2, . . . )

is infinite anda has no oldest ancestor and, therefore,a ∈ A∞. Thus,B∞ ⊆ f (A∞).

6. Thus, f (A∞) = B∞ and f : A∞ → B∞ is a surjection, and therefore, abijection.

7. The proof forg is the same.

�

• There must be an infinite number of infinite cardinals.

• Corollary 6.2.2 (0,1) ≈ [0,1]

Proof: Easy.�

• Any two intervals of real numbers have the same cardinality.

• The cardinality ofP(Z+) is c. That is

|P(Z+)| = |R| .

• For any countably infinite setX

|P(X)| = |R| .

• Corollary 6.2.3 P(Z+) ≈ RProof:

1. It suffices to showP(Z+) ≈ [0,1).

2. Let f : P(Z+)→ [0,1) be a function defined for a setA ⊂ Z+ by

f (A) = 0.a1a2 . . .

(a decimal expansion), whereai = 0 if i ∈ A andai = 1 if i < A.

3. Claim: f is an injection.

4. So,P(Z+) � [0,1).


5. Letg : [0,1)→ P(Z+) be a function defined by

g(0.a1a2 . . . ) = {i ∈ Z+|ai = 0}

(a binary expansion).

6. Claim:g is an injection.

7. So, [0,1) � P(Z+).

8. ThusP(Z+) ≈ [0,1).

�

6.2.5 Homework




6.3. LECTURE 21. COLLECTIONS OF SETS 93

6.3 LECTURE 21. Collections of Sets

6.3.1 Russell’s Paradox

• A collection of sets{Ai}i∈I is countable collectionif the indexing setI iscountable anduncountable if the indexing set is uncountable.

• Proposition. There is no universal set of all sets.Proof: Let U be a universal set of all sets. ThenU ≺ P(U). Therefore,Udoes not contain its own power setP(U), which is a contradiction.

• There are logical problems with assuming that any combination of sets re-sults in another set.

• Let A = {A|A < A}, that is the set of all sets that are not elements ofthemselves.

• Russell’s Paradox. If A ∈ A, then, by its definition,A < A. Also, ifA < A, then, by its definition,A ∈ A.

• Not every collection of sets can be considered a set!

• Axioms of set theory: if all the sets are subsets of a universal set, then theoperations of union, intersection and Cartesian product generate sets.

6.3.2 Countable Unions of Countable Sets

• A finite union of countable sets is countable.

• Theorem 6.3.1 Let {Ai}i∈Z+be a countably infinite collection of count-

ably infinite sets. Then⋃

i∈Z+Ai is also a countably infinite set.

Proof: Diagonal argument.�

• An uncountable union of countable sets may be uncountable.

• Example.

Definition 6.3.1 Let {Ai}i∈Z+be a countably infinite collection of sets.

TheCartesian product of the collection�

i∈Z+Ai is the set of all se-

quences(a1,a2, . . . ) such thatai ∈ Ai for all i ∈ Z+.


• Example

•Theorem 6.3.2 Let {Ai}i∈Z+

be a countably infinite collection of sets.If each Ai contains at least two elements, then the Cartesian product�

i∈Z+Ai of the collection is uncountable.


1. Suppose thatA =�

i∈Z+Ai is countable.

2. There is a bijectionf : Z+ → A.

3. Let f (i) = Ai.

4. Ai are sequencesAi = (ai1,a

i2,a

i3, . . . ), whereai

j ∈ Aj.

5. Choosebi ∈ Ai , aii.

6. LetB = (b1,b2, . . . ).

7. ThenB ∈ A, butB , Ai for any i ∈ Z+. (Contradiction).

�

•

Definition 6.3.2 LetΛ be an indexing set and let{Aλ}λ∈Λ be a collec-tion of sets. TheCartesian product

�λ∈Λ Aλ of the collection{Aλ}λ∈Λ

is the set of all functions

f : Λ→ ⋃λ∈Λ Aλ

such thatf (λ) ∈ Aλ for all λ ∈ Λ.

• Example.

• Suppose that all setsAλ are nonempty, i.e.Aλ , ∅ for all λ ∈ Λ.

• Question: Is the Cartesian product�

λ∈Λ Aλ nonempty?

• Answer: Not trivial.

• Foundation of set theory:Zermelo-Frankel Axioms.

• ZF (Zermelo-Frankel Set Theory): existence of the empty set, existenceof the power set of a set, unions, intersections etc.

• The Axiom of Choice:


6.3. LECTURE 21. COLLECTIONS OF SETS 95

If Λ , ∅ andAλ , ∅ for all λ ∈ Λ, then�

λ∈Λ Aλ , ∅.

cannot be proven or disproven from ZF.

• The systemZFC is the set theory based on the Zermelo-Frankel axiomstogether with the Axiom of Choice.

• ZFC is needed to prove that the cardinal numbers are linearly ordered.

If A, B are sets, then eitherA � B or B � A.

6.3.3 Homework



Chapter 7

The Real and Complex Numbers

7.1 LECTURE 22. Fields

7.1.1 Fields

•

Definition 7.1.1 A field F is a nonempty set with two binary oper-ations: addition, +, and multiplication , ·, that satisfy the followingaxioms:

1. Associativity of addition:

(x + y) + z = x + (y + z), ∀x, y, z ∈ F

2. Commutativity of addition :

x + y = y + x, ∀x, y ∈ F

3. Additive identity (or zero element, or zero):

∃ 0 ∈ F such that x + 0 = x, ∀x ∈ F

4. Additive inverse (or negative):

∀x ∈ F, ∃(−x) ∈ F such that x + (−x) = 0

97

98 CHAPTER 7. THE REAL AND COMPLEX NUMBERS

Definition 7.1.2

1. Associativity of multiplication :

(x · y) · z = x · (y · z), ∀x, y, z ∈ F

2. Commutativity of multiplication :

x · y = y · x, ∀x, y ∈ F

3. Multiplicative identity (or one):

∃ 1 ∈ F such that x · 1 = x, ∀x ∈ F, and 1 , 0

4. Multiplicative inverse (or the inverse):

∀x ∈ F, x , 0, ∃x−1 ∈ F such that x · x−1 = 1

5. Distributivity :

x · (y + z) = x · y + x · z, ∀x, y, z ∈ F

• Z is not a field.

• Q, R, C are fields.

7.1.2 Elementary Properties of Fields

•

Proposition 7.1.1 Let F be a field anda,b, c ∈ F. Then

1. a0 = 0

2. (−a)b = a(−b) = −(ab)

3. −(−a) = a

4. (−a)(−b) = ab

5. a(b− c) = ab− bc

Proof: Exercise.


7.1. LECTURE 22. FIELDS 99

�

Proposition 7.1.2

1. (−1)a = −a

2. (−1)(−1) = 1

3. If a + b = a + c, thenb = c

4. If a , 0 andab = ac, thenb = c

5. If a , 0, then(a−1)−1 = a

Proof: Exercise.�

7.1.3 Ordered Fields

•

Definition 7.1.3 LetF be a field. We sayF is anordered field if thereexists a subsetP of F satisfying two properties:

1. P is closed with respect to addition and multiplication, that is

if x, y ∈ P, thenx + y ∈ P andxy ∈ P.

2. For everyx ∈ F exactly one of the following is true:

x ∈ P, −x ∈ P, or x = 0.

The subsetP is the set ofpositiveelements ofF.

•Definition 7.1.4 Let F be an ordered field andP be the set of positiveelements ofF. Let x, y ∈ F. Then

x < y (or y > x) if y− x ∈ P.

Notation.

x < y is read “x is less thany” (or “ y is greater than x”)

if x < y, theny > x

if x < y or x = y, thenx ≤ y

if x > y or x = y, thenx ≥ y

if x ≤ y, theny ≥ x

• If x ∈ P, thenx > 0.

•

Proposition 7.1.3 Let F be an ordered field andP be a subset of posi-tive elements ofF. Leta,b, c ∈ F.

1. Exactly one of the following holds:

a < b, b < a or b = a.

2. If a > 0, then−a < 0. (If a < 0, then−a > 0.)

3. If a > 0 andb > 0, thena + b > 0 andab> 0.

4. If a > 0 andb < 0, thenab< 0.

5. If a < 0 andb < 0, thenab> 0.

6. If a 0, thenac< bc.

9. If a bc.

10. If a , 0, thena2 ∈ P.

Proof: Exercise.�

• The set of positive elementsP of any fieldF does not have a smallest ele-ment.

•

Theorem 7.1.1 LetF be an ordered field andP be a subset of positiveelements ofF. Then:

1. 1 ∈ P.

2. If x ∈ P, thenx−1 ∈ P.

3. P is an infinite set.

Proof:

1. (1.) Suppose 1< P. Then (−1) ∈ P and 1= (−1)(−1) ∈ P, which is acontradiction.

2. (2.) Exercise.

3. (3.) Letn ∈ Z+ andxn = 1 + · · · + 1︸︷︷︸n

.

4. Claim: xn , xm for n , m.

5. Supposexn = xm for somen > m. Then 0= xn− xm = 1 + · · · + 1︸︷︷︸n−m

∈ P,

which is a contradiction.

�

• Corollary 7.1.1 Every ordered field is infinite.

7.1.4 Finite Fields

• Recall that for anyn ∈ Z+, n > 1, Zn = {[0], [1], . . . , [n− 1]} is the set ofequivalence classes modulon.

• Theorem 7.1.2 Let n ∈ Z+, n > 1. ThenZn is a field if and only ifnis prime.

Proof:

1. (I). Letn be prime.

2. Zn satisfies all field axioms except possibly the existence of multiplica-tive inverses.

3. Let [a] ∈ Zn, [a] , [0].

4. Then [a] = [1], . . . , [n− 1].

5. Thenn does not dividea since 0< a < n.

6. Sincen is prime, gcd(a,n) = 1 (a andn are relatively prime).

7. So,∃x, y ∈ Z such thatax+ ny = 1.

8. Thus, [ax+ ny] = [1], and since [n] = [0], then [a][ x] = [1].

9. Hence, [x] = [a]−1 (multiplicative inverse).

10. So, ifn is prime, thenZn is a field.

11. (II). Conversely (claim): ifZn is a field, thenn is prime.

12. Supposen is not prime.

13. There area,b ∈ Z such that 1< a < n and 1< b < n andn = ab.

14. Then [a][b] = [0], but [a] , [0] and [b] , [0].

15. Claim: [a] does not have multiplicative inverse.

16. Suppose∃x ∈ Z such that [a][ x] = [1].

17. Since [a][b] = [0], then [0]= [b].

18. However, [b] , [0] since 1< b < n.

19. Contradiction. So, [a] does not have a multiplicative inverse andZn isnot a field.

�

• For a primep the finite fieldZp is not ordered.

• Algebraic number fields.

• Example.Q

(√2)

={r + s

√2|r, s ∈ Q

}

• Galois theory. Idea: Adding solutions of algebraic equations to a field.

• Let p be a prime number,ak ∈ Zp, and

f (x) = anxn + · · · a1x + a0

be anirreducible (cannot be factorized) polynomial of a formal variablexwith coefficientsak in Zp.


• Let α be a solution of the equation

f (α) = [0] .

• ThenZp(α) = {x + yα|x, y ∈ Zp}

is a finite field and there ism ∈ Z+ such that

|Zp(α)| = pm .

7.1.5 Homework

• Reading: Sect. 7.1.

• Exercises: 7.1[2,6,8,20]



7.2 LECTURE 23. The Real Numbers

7.2.1 Bounded Sets

•

Definition 7.2.1 LetS be a nonempty set of real numbers.

1. The setS is bounded aboveif there exists a real numberx ∈ Rsuch that

a ≤ x for all a ∈ S.

Such a numberx is called anupper bound for the setS.

2. The setS is bounded belowif there exists a real numbery ∈ Rsuch that

a ≥ y for all a ∈ S.

Such a numbery is called alower bound for the setS.

3. The setS is bounded if it is bounded below and bounded above.

• Examples.

7.2.2 Least Upper Bound and Greatest Lower Bound

• Example

•

Definition 7.2.2 Let S be a nonempty set of real numbers boundedabove. Thenx ∈ R is theleast upper bound(or supremum) of S if:

1. x is an upper bound ofS, and

2. if z is an upper bound ofS, thenx ≤ z.

•

Definition 7.2.3 Let S be a nonempty set of real numbers boundedbelow. Theny ∈ R is thegreatest lower bound(or infimum ) of S if:

1. y is a lower bound ofS, and

2. if z is a lower bound ofS, theny ≥ z.


7.2. LECTURE 23. THE REAL NUMBERS 105

• Notation:

y = inf S, x = supS

• Examples.

• Definition 7.2.4 Least Upper Bound Axiom. Any nonempty set ofreal numbers that is bounded above has a least upper bound.

• Any nonempty set ofR bounded below has a greatest lower bound.

• If a supremum of a set exists, then it is unique.

• If a infimum of a set exists, then it is unique.

7.2.3 The Archimedean Principle

• Theorem 7.2.1 Let a,b ∈ R+ be positive real numbers. Then thereexists a positive integern ∈ Z+ such thatna> b.


1. Supposena≤ b, ∀n ∈ Z+.

2. Letz = sup{na | n ∈ Z+}.3. ∃x ∈ S such thatz− a < x.

4. ∃m ∈ Z+ such thatx = ma.

5. Thenz< (m+ 1)a and (m+ 1)a ∈ S (contradiction).

�

• Corollary 7.2.1 Letb ∈ R+ be a positive real number. Then there existsa positive integern ∈ Z+ such that1n < b.

Proof: ∃ n ∈ Z+ such thatn > 1b, or 1

n < b.�

• Lemma 7.2.1 Let x ∈ R. Then there exists an integerm ∈ Z such thatm− 1 ≤ x < m.

Proof:

1. Let x > 0.

2. ∃n ∈ Z+ such thatn > x.

3. ∃ a smallest such integer, call itm.

4. Thenx < m andm− 1 ≥ x.

5. If x = 0, thenm = 1.

6. Let x < 0.

7. Then (−x) > 0.

8. ∃n ∈ Z+ such thatn > (−x).

9. Thenn + x > 0.

10. Thus,∃k ∈ Z+ such thatk− 1 ≤ n + x < k.

11. Therefore,m− 1 ≤ x < m, wherem = k− n.

�

• Theorem 7.2.2 Between any two distinct real numbers there is a ratio-nal number.

Proof:

1. Let x, y ∈ R such thatx < y.

2. Theny− x > 0.

3. So,∃n ∈ Z+ such thatn > 1y−x.

4. Thus,n(y− x) > 1, ornx+ 1 < ny.

5. ∃m ∈ Z+ such thatm− 1 ≤ nx< m.

6. So,nx< m.

7. Finally,m≤ nx+ 1 < ny, or m< ny.

8. Thusnx< m< nyandx < mn < y.

�

7.2.4 Incompleteness of Rationals

• The set of rationalsQ does not satisfy the Least Upper Bound Axiom.

• Example. The set of rational numbersS = {r ∈ Q | r2 < 2} does not have asupremum.

7.2. LECTURE 23. THE REAL NUMBERS 107

• Theorem 7.2.3 Let x ∈ R+ be a positive real number. Then thereexists a positive real numbery ∈ R+ such thaty2 = x.

Proof:

1. LetS = {t ∈ R+ | t2 < x}2. Claim:S , ∅.3. Let t = x

x+1.

4. Thent ∈ S.

5. Claim:S is bounded above.

6. x + 1 is an upper bound ofS.

7. Lety = supS.

8. Theny > 0.

9. Claim:y2 = x.

10. Supposey2 < x.

11. ∃n ∈ Z+ such that1n <x−y2

2y+1.

12. Then (y + 1n)2 < x.

13. So,y + 1n ∈ S (contradiction).

14. Supposey2 > x.

15. ∃m ∈ Z+ such that1m < y2−x2y .

16. Then (y− 1m)2 > x.

17. Leta ∈ S. Thena2 < x. Thusa2 < (y− 1m)2.

18. Thena < y− 1m.

19. So,y− 1m is an upper bound forS buty− 1

m < y (Contradiction).

20. Therefore,y2 = x.

�

• The square root of a positive real number is unique.

7.2.5 Homework


• Exercises: 7.2[2,7,9,15 ]


7.3 LECTURE 24. The Complex Numbers

7.3.1 Complex Numbers

•

Definition 7.3.1 The set ofcomplex numbers is the set of all or-dered pairs(a,b) of real numbers together with two binary operations,addition,+ and multiplication,·, defined by

1. (a,b) + (c,d) = (a + c,b + d)

2. (a,b) · (c,d) = (ac− bd,ad+ bc)

• Notation. The set of complex numbers is denoted byC

• The set of complex numbers forms a field.

1. Additive identity (zero): (0,0)

2. Additive inverse:−(a,b) = (−a,−b)

3. Multiplicative identity: (1,0)

4. Multiplicative inverse:

(a,b)−1 =

(a

a2 + b2,−b

a2 + b2

)

• Theorem 7.3.1 The set of complex numbersC is a field with respectto the binary operation of addition and multiplication.

• Two complex numbers (a,b) and (c,d) are equal if and only ifa = b andc = d.

• The setC is not an ordered field.

• The complex numbers of the form (a,0) can be identified with the subset ofreal numbersR.

• R is a subfield ofC

• Notation. Imaginary identity

i = (0,1)


7.3. LECTURE 24. THE COMPLEX NUMBERS 109

• Every complex number (a,b) can be written as

z = (a,b) = a(1,0) + b(0,1) = a + ib

• The square of the imaginary identity

i2 = (−1,0) = −1

• If z = a + ib, thena = Rez is thereal part of z.

• If z = a + ib, thenb = Im z is theimaginary part of z.

• The setC of complex numbers is the Cartesian productR×R. The planeR2

is called thecomplex plane.

7.3.2 Conjugation and Absolute Value

•

Definition 7.3.2 Letz = a+ib ∈ C be a complex number. The complexnumber

z = a− ib

is thecomplex conjugateof z.The real number

|z| =√

a2 + b2

is theabsolute valueof z.

• Geometric interpretation in the complex plane.

•

Lemma 7.3.1 Let z,w ∈ C. Then

1. z+ w = z+ w

2. zw= zw

3. zz = |z|2

4. z+ z = 2 Rez

5. z− z = 2i Im z

6. If z, 0, thenz−1 = z|z|2



7.3.3 Solutions of Equations

•

Theorem 7.3.2 Let n ∈ Z+, n ≥ 1, be a positive integer greater than1, a j ∈ R, j = 0,1,2, . . . , n, be real numbers such thatan , 0, and

p(z) = anzn + · · · + a1z+ a0

be a polynomial of degreen. Then, if p(z) = 0 for somez ∈ C, thenp(z) = 0.(That is, ifz is a root of the polynomial, then the complex conjugatez isalso a root.)

Proof: Easy.

7.3.4 Polar Form

• Let z = x + iy. Define

r = |z| =√

x2 + y2 and tanθ =yx.

so that

x = r cosθ, y = r sinθ

Thepolar form of z is

z = r(cosθ + i sinθ)

• Argument of z is the angleθ = argz.

• Non-uniqueness.

• arg 0 is not defined.

•

Theorem 7.3.3 Let z,w ∈ C. If

z = r(cosθ + i sinθ) andw = s(cosφ + i sinφ),

then

zw= rs[cos(θ + φ) + i sin(θ + φ)]

Proof: Use trigonometric identities.


7.3. LECTURE 24. THE COMPLEX NUMBERS 111

•

Theorem 7.3.4 DeMoivre’s Theorem.Let z ∈ C andn ∈ Z+. If

z = r(cosθ + i sinθ),

then

zn = rn[cos(nθ) + i sin(nθ)]


• Define the complex exponential function

eiθ = cosθ + i sinθ .

7.3.5 Complex Roots

•Definition 7.3.3 Let z ∈ C andn ∈ Z+. Annth root of z is a complexnumberw such that

wn = z.

•

Theorem 7.3.5 Let z ∈ C andn ∈ Z+. If

z = r(cosθ + i sinθ)

andr , 0, thenz has exactlyn n-th roots given by

wk = r1n

[cos

(θ + 2πk

n

)+ i sin

(θ + 2πk

n

)]

for k = 0,1,2, . . . , n− 1.

Proof: Calculation.

•Definition 7.3.4 Let n ∈ Z+. An nth root of unity is a complexnumberw such that

wn = 1.



•

Theorem 7.3.6 Let n ∈ Z+. Then there are exactlyn n-th roots ofunity given by

wk = cos

(2πkn

)+ i sin

(2πkn

)

for k = 0,1,2, . . . , n− 1.

Proof: Easy.

•

Corollary 7.3.1 Letn ∈ Z+ and

ω = cos(

2πn

)+ i sin

(2πn

).

Then then-th roots of unity are

wk = ωk

for k = 0,1,2, . . . , n− 1.

7.3.6 Homework


• Exercises: 7.3[6,10,13]


Bibliography

[1] R. J. Bond and W. J. Keane,An Introduction to Abstract Mathematics,Brooks/Cole, 1999

[2] J. H. Goodfriend,Gateway to Higher Mathematics, Jones & Bartlett, 2005

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